Projectile Motion Practice Questions — JAMB UTME

Question 1Question →

A body is projected horizontally from the top of a cliff $45\text{ m}$ high. If it lands on flat ground at a horizontal distance of $120\text{ m}$ from the base of the cliff, what is the speed of the body just before it strikes the ground? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 50

Answer

The speed of the body just before striking the ground is 50 m/s.

The time of fall is determined by the height of

45\text{ m}

, yielding

t = \sqrt{2h/g} = 3\text{ s}

. The horizontal speed is constant at

120 / 3 = 40\text{ m/s}

. The vertical velocity gained on impact is

v_y = gt = 30\text{ m/s}

. Combining these mutually perpendicular velocity components gives a final impact speed of

v = \sqrt{40^2 + 30^2} = 50\text{ m/s}

.

Step-by-Step Solution

1

Calculate the time of flight from the vertical height

t = 3 s

Vertical acceleration is constant under gravity while initial vertical velocity is zero.

2

Compute the constant horizontal component of velocity

v_x = 40 m/s

Horizontal speed is uniform because zero horizontal force acts on the projectile.

3

Compute the final vertical component of velocity at impact

v_y = 30 m/s

Vertical speed increases linearly with time due to gravitational acceleration.

4

Determine the magnitude of the resultant velocity vector

v = 50 m/s

The horizontal and vertical components are perpendicular, so their vector sum uses the Pythagorean theorem.

Key Concept

Horizontal Projection and Impact Velocity Vector

Estimated Time:1m 30s

Question 2Question →

An athlete throws a javelin from ground level such that its initial vertical component of velocity is $40\text{ m/s}$ and its initial horizontal component of velocity is $30\text{ m/s}$ . What is the horizontal distance in metres covered by the javelin when it reaches a height of $35\text{ m}$ above the ground for the first time? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 30

Answer

The horizontal distance covered by the javelin when it reaches a height of 35 m for the first time is 30 m.

Using

y = u_y t - \frac{1}{2}g t^2

with

u_y = 40\text{ m/s}

,

y = 35\text{ m}

, and

g = 10\text{ m/s}^2

yields

35 = 40t - 5t^2

. Dividing by 5 gives

t^2 - 8t + 7 = 0

, which factors to

(t - 1)(t - 7) = 0

. The roots are

t = 1\text{ s}

(ascent) and

t = 7\text{ s}

(descent). For the first time,

t = 1\text{ s}

. The horizontal displacement is

x = u_x t = 30\text{ m/s} \times 1\text{ s} = 30\text{ m}

.

Step-by-Step Solution

1

Set up the vertical motion equation to find the time when height is 35 m

35 = 40t - 5t^2

Vertical displacement in projectile motion depends on the vertical initial velocity component and acceleration due to gravity.

2

Solve the quadratic equation for time

t

t^2 - 8t + 7 = 0 \implies t = 1\text{ s} \text{ or } t = 7\text{ s}

A projectile reaches a given non-peak height twice: once ascending and once descending.

3

Select the first time value and calculate horizontal distance

x = u_x \times t = 30 \times 1 = 30\text{ m}

Horizontal velocity remains constant throughout the flight, so distance is speed multiplied by time.

Key Concept

Independence of vertical and horizontal components in projectile motion

Question 3Question →

An object is projected from ground level at an angle of $60^\circ$ to the horizontal. If the horizontal component of its initial velocity is $25\text{ m/s}$ , calculate the maximum height reached by the object in meters. (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 93.75

Answer

The maximum height reached by the object is

93.75\text{ m}

.

The maximum vertical height attained by a projectile depends on its vertical velocity component

u_y = u \sin \theta

. Resolving the initial velocity gives

u = 50\text{ m/s}

and

u_y = 25\sqrt{3}\text{ m/s}

. Substituting into

H = \frac{u_y^2}{2g}

yields

93.75\text{ m}

.

Step-by-Step Solution

1

Find the magnitude of the initial velocity

u = 50\text{ m/s}

The horizontal velocity component remains constant throughout flight and is given by

u_x = u \cos \theta

.

2

Calculate the initial vertical velocity component

u_y = 25\sqrt{3}\text{ m/s}

Vertical component of velocity is calculated using

u_y = u \sin \theta

.

3

Calculate the maximum height

H = 93.75\text{ m}

At maximum height, vertical velocity is zero, giving

H = \frac{u_y^2}{2g}

.

Key Concept

Resolution of velocity components in projectile motion and calculation of maximum height

Question 4Question →

A projectile is launched from ground level over flat terrain. At time $t = 2\text{ s}$ after launch, the projectile passes through a point located $60\text{ m}$ horizontally and $60\text{ m}$ vertically from its launch point. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the total horizontal range of the projectile in meters?

Show answer & explanation

Answer: 240

Answer

The total horizontal range of the projectile is

240\text{ m}

.

The horizontal motion occurs at a constant velocity of

30\text{ m/s}

calculated from

\frac{60\text{ m}}{2\text{ s}}

. Substituting the vertical position (

60\text{ m}

) and time (

2\text{ s}

) into

y = u_y t - 5t^2

yields an initial vertical velocity of

40\text{ m/s}

. The total duration in the air is

T = \frac{2(40)}{10} = 8\text{ s}

. The total horizontal range is therefore

30\text{ m/s} \times 8\text{ s} = 240\text{ m}

.

Step-by-Step Solution

1

Determine the horizontal component of velocity

v_x = 30\text{ m/s}

Horizontal velocity remains constant throughout flight because there is no horizontal acceleration.

2

Determine the initial vertical component of velocity

u_y = 40\text{ m/s}

Applying the vertical displacement equation

y = u_y t - \frac{1}{2}g t^2

with

y = 60\text{ m}

,

t = 2\text{ s}

, and

g = 10\text{ m/s}^2

.

3

Calculate the total time of flight

T = 8\text{ s}

The projectile completes its full parabolic trajectory when vertical displacement returns to zero, given by

T = \frac{2 u_y}{g}

.

4

Calculate the total horizontal range

R = 240\text{ m}

The total range is the product of the constant horizontal velocity component and total time of flight (

R = v_x \times T

).

Key Concept

Independence of horizontal and vertical components of projectile motion

Question 5Question →

A stone is projected from ground level with an initial velocity of $40\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. What is the total time of flight of the stone, in seconds? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 4

Answer

The total time of flight of the stone is

4\text{ s}

.

The total time of flight

T

for a projectile launched over level ground is calculated using

T = \frac{2 u \sin \theta}{g}

. Substituting

u = 40\text{ m/s}

,

\theta = 30^\circ

, and

g = 10\text{ m/s}^2

yields

T = \frac{2 \times 40 \times 0.5}{10} = 4\text{ s}

.

Step-by-Step Solution

1

Find the vertical component of the launch velocity

u_y = 20\text{ m/s}

The vertical motion determines the time the projectile remains in the air.

2

Calculate the total time of flight

T = 4\text{ s}

Applying

T = \frac{2 u \sin \theta}{g} = \frac{2 \times 20}{10} = 4\text{ s}

gives the total duration before landing back at ground level.

Key Concept

Time of Flight in Projectile Motion

Estimated Time:45s

Question 6Question →

A projectile is launched from ground level with a horizontal velocity component of $15\text{ m/s}$ and a vertical velocity component of $20\text{ m/s}$ . Neglecting air resistance, what is the magnitude of the velocity of the projectile at its maximum height?

15\text{ m/s}

0\text{ m/s}

20\text{ m/s}

25\text{ m/s}

Show answer & explanation

Answer: $15\text{ m/s}$

Answer

The magnitude of the velocity of the projectile at its maximum height is

15\text{ m/s}

.

In projectile motion under gravity without air resistance, the horizontal component of velocity remains constant throughout flight (

v_x = 15\text{ m/s}

). At the highest point (apex), the vertical component of velocity momentarily becomes zero (

v_y = 0\text{ m/s}

). Therefore, the magnitude of the velocity at the maximum height is equal to the horizontal component, which is

15\text{ m/s}

.

Step-by-Step Solution

1

Identify the velocity components at the maximum height of a projectile

Vertical component

v_y = 0\text{ m/s}

and horizontal component

v_x = u_x = 15\text{ m/s}

.

Gravity acts vertically, reducing vertical velocity to zero at the peak, while horizontal velocity remains constant in the absence of air resistance.

2

Calculate the total magnitude of velocity at maximum height

v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 0^2} = 15\text{ m/s}

.

The magnitude of the resultant velocity vector is derived using the Pythagorean theorem.

Key Concept

Velocity at Maximum Height in Projectile Motion

Question 7Question →

A projectile is launched from ground level over flat terrain with a constant horizontal velocity component of $10\text{ m/s}$ . At a height of $40\text{ m}$ above the ground, the magnitude of its vertical velocity component is equal to its horizontal velocity component. Taking $g = 10\text{ m/s}^2$ , calculate the maximum height reached by the projectile in meters.

Show answer & explanation

Answer: 45

Answer

The maximum height reached by the projectile is

45\text{ m}

.

At the height of

40\text{ m}

, the vertical velocity is equal to the horizontal velocity of

10\text{ m/s}

. Applying the vertical kinematic relation

v_y^2 = u_y^2 - 2gh

yields

10^2 = u_y^2 - 2(10)(40)

, which gives

u_y^2 = 900\text{ m}^2/\text{s}^2

. The maximum height attained above ground level occurs when

v_y = 0

, calculated as

H_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}

.

Step-by-Step Solution

1

Identify the vertical component of velocity at the given height

v_y = 10\text{ m/s}

at

h = 40\text{ m}

The problem states that at

h = 40\text{ m}

, the vertical velocity component equals the constant horizontal component

u_x = 10\text{ m/s}

.

2

Determine the initial vertical launch velocity component

u_y

u_y^2 = 900\text{ m}^2/\text{s}^2 \implies u_y = 30\text{ m/s}

Applying the vertical motion equation

v_y^2 = u_y^2 - 2gh

gives

10^2 = u_y^2 - 2(10)(40) \implies u_y^2 = 100 + 800 = 900

.

3

Calculate the maximum height

H_{\text{max}}

H_{\text{max}} = 45\text{ m}

At maximum height, the vertical velocity becomes zero. Using

H_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}

.

Key Concept

Vertical Kinematics and Maximum Height of a Projectile

Question 8Question →

A particle is projected from horizontal ground at an initial angle $\theta$ to the horizontal. At its maximum height $H = 20\text{ m}$ , its kinetic energy is exactly half of its initial kinetic energy at launch. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the horizontal range of the projectile?

40\text{ m}

80\text{ m}

20\text{ m}

0\text{ m}

Show answer & explanation

Answer: $80\text{ m}$

Answer

The horizontal range of the projectile is

80\text{ m}

.

At the trajectory's highest point, the vertical component of velocity becomes zero while the horizontal component

u\cos\theta

remains unchanged. The kinetic energy at the apex is therefore

\frac{1}{2}m(u\cos\theta)^2 = E_{k,i}\cos^2\theta

. Setting

\cos^2\theta = \frac{1}{2}

gives

\theta = 45^\circ

. For a

45^\circ

launch angle, the horizontal range

R = \frac{u^2}{g}

is related to the maximum height

H = \frac{u^2}{4g}

by

R = 4H

. Substituting

H = 20\text{ m}

yields

R = 80\text{ m}

.

Step-by-Step Solution

1

Relate kinetic energy at maximum height to initial kinetic energy.

Initial kinetic energy

E_k(0) = \frac{1}{2}m u^2

. At peak height, vertical velocity component

v_y = 0

, so velocity is

v_x = u\cos\theta

. Thus,

E_k(\text{peak}) = \frac{1}{2}m (u\cos\theta)^2 = E_k(0)\cos^2\theta

.

At maximum height, only the horizontal component of velocity remains.

2

Calculate the launch angle

\theta

.

Given

E_k(\text{peak}) = \frac{1}{2} E_k(0)

, we have

\cos^2\theta = \frac{1}{2}

, giving

\theta = 45^\circ

.

Setting the energy expression equal to the given condition enables solving for the angle.

3

Relate maximum height

H

to horizontal range

R

for

\theta = 45^\circ

.

For

\theta = 45^\circ

,

H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2}{4g}

and

R = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g}

. Therefore,

R = 4H

.

Standard formulas for maximum height and range express both quantities in terms of launch speed

u

and acceleration due to gravity

g

.

4

Substitute the maximum height value into the range formula.

R = 4 \times 20\text{ m} = 80\text{ m}

.

Multiplying the given peak height of

20\text{ m}

by

4

yields the exact horizontal range.

Key Concept

Kinetic Energy Conservation and Range-Height Relationship in Projectile Motion

Estimated Time:2m 0s

Question 9Question →

A cannonball is fired from ground level with an initial speed of $50\text{ m/s}$ at an angle $\theta$ above the horizontal such that $\tan \theta = \frac{4}{3}$ . Assuming acceleration due to gravity $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the magnitude of the velocity of the cannonball at the highest point of its trajectory?

30\text{ m/s}

0\text{ m/s}

40\text{ m/s}

50\text{ m/s}

Show answer & explanation

Answer: $30\text{ m/s}$

Answer

30\text{ m/s}

At the maximum height of a projectile's flight, the vertical velocity component drops to zero due to gravity, while the horizontal component remains unchanged because horizontal acceleration is zero. For a launch speed of

50\text{ m/s}

at an angle with

\tan \theta = \frac{4}{3}

,

\cos \theta = \frac{3}{5}

. The magnitude of the velocity at apex is therefore equal to

u_x = 50 \times \frac{3}{5} = 30\text{ m/s}

.

Step-by-Step Solution

1

Resolve the initial velocity vector into orthogonal horizontal and vertical components.

Given

\tan \theta = \frac{4}{3}

, the trigonometric ratios are

\cos \theta = \frac{3}{5}

and

\sin \theta = \frac{4}{5}

. Thus,

u_x = 50 \times \frac{3}{5} = 30\text{ m/s}

and

u_y = 50 \times \frac{4}{5} = 40\text{ m/s}

.

Resolving 2D projectile motion into independent orthogonal components simplifies kinematic analysis.

2

Determine the velocity components at the apex (highest point) of the trajectory.

At maximum height, the vertical velocity component

v_y = 0\text{ m/s}

. Since there is no horizontal acceleration, the horizontal velocity component remains

v_x = u_x = 30\text{ m/s}

.

Gravity acts purely vertically, reducing vertical velocity to zero at the apex while leaving horizontal velocity unchanged.

3

Calculate the magnitude of total velocity at maximum height.

v = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 0^2} = 30\text{ m/s}

.

The magnitude of a 2D velocity vector is determined by combining its orthogonal components using the Pythagorean formula.

Key Concept

Velocity components at maximum height in projectile motion

Question 10Question →

A projectile is launched from horizontal ground with an initial speed of $60\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. Taking $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the speed of the projectile at its maximum height?

0\text{ m/s}

30\text{ m/s}

30\sqrt{3}\text{ m/s}

60\text{ m/s}

Show answer & explanation

Answer: $30\text{ m/s}$

Answer

The speed of the projectile at its maximum height is

30\text{ m/s}

.

In two-dimensional projectile motion, gravity acts purely in the vertical direction. At the apex (maximum height), the vertical velocity component

v_y

drops to zero. However, the horizontal velocity component

v_x = u \cos \theta

remains constant throughout the entire flight because air resistance is neglected. Consequently, the speed at maximum height equals

u \cos 60^\circ = 60 \times 0.5 = 30\text{ m/s}

.

Step-by-Step Solution

1

Resolve initial launch velocity into orthogonal horizontal and vertical components.

u_x = u \cos \theta = 60 \cos 60^\circ = 30\text{ m/s}

and

u_y = u \sin \theta = 60 \sin 60^\circ = 30\sqrt{3}\text{ m/s}

.

Projectile motion consists of independent horizontal and vertical motions.

2

Determine the velocity components at the apex (maximum height).

At the peak, vertical velocity

v_y = 0\text{ m/s}

while horizontal velocity remains

v_x = u_x = 30\text{ m/s}

.

Gravity acts vertically causing

v_y

to become zero at peak height, whereas no horizontal force acts, leaving

v_x

unchanged.

3

Calculate the magnitude of total velocity (speed) at maximum height.

v = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 0^2} = 30\text{ m/s}

.

Speed is the resultant magnitude of orthogonal velocity components.

Key Concept

Velocity at Maximum Height in Projectile Motion

Question 11Question →

A stone is thrown into the air with an initial velocity of $20\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. What is the magnitude of its velocity at the highest point of its trajectory?

0\text{ m/s}

10\text{ m/s}

17.3\text{ m/s}

20\text{ m/s}

Show answer & explanation

Answer: $10\text{ m/s}$

Answer

The magnitude of the velocity at the highest point is

10\text{ m/s}

.

At the peak of a projectile's trajectory, the vertical velocity component vanishes (

v_y = 0\text{ m/s}

), but horizontal motion continues at a constant speed (

v_x = u \cos \theta

). Substituting

u = 20\text{ m/s}

and

\theta = 60^\circ

yields

v_x = 20 \times 0.5 = 10\text{ m/s}

.

Step-by-Step Solution

1

Identify the velocity components at the apex

At maximum height, the vertical velocity component is

v_y = 0\text{ m/s}

, while the horizontal component remains constant at

v_x = u_x = u \cos \theta

.

Horizontal acceleration is zero when air resistance is neglected.

2

Calculate the horizontal component of velocity

v_x = 20 \times \cos(60^\circ) = 20 \times 0.5 = 10\text{ m/s}

.

The trigonometric cosine function gives the horizontal projection of the initial velocity vector.

3

Determine total velocity magnitude at the apex

v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 0^2} = 10\text{ m/s}

.

Since the vertical component is zero, the total velocity at the peak equals the horizontal component.

Key Concept

Velocity components at maximum height in projectile motion

Question 12Question →

A body is projected from ground level with an initial speed of $40\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Calculate the time taken, in seconds, for the body to reach its maximum height. (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 2

Answer

The time taken to reach maximum height is

2\text{ s}

.

The initial vertical velocity component is

u_y = u \sin(30^\circ) = 40 \times 0.5 = 20\text{ m/s}

. Under gravitational deceleration (

g = 10\text{ m/s}^2

), the vertical speed drops to zero at maximum height after

t = \frac{u_y}{g} = \frac{20}{10} = 2\text{ s}

.

Step-by-Step Solution

1

Find the vertical component of initial velocity (

u_y

)

u_y = 40 \sin(30^\circ) = 20\text{ m/s}

Only the vertical component of initial velocity determines the time to reach maximum height.

2

Calculate the time to maximum height (

t

)

t = \frac{u_y}{g} = \frac{20}{10} = 2\text{ s}

At maximum height, vertical velocity

v_y = 0

, giving

t = \frac{u_y}{g}

.

Key Concept

Time to reach maximum height in projectile motion

Question 13Question →

A cannonball is fired from level ground with an initial speed of $20\text{ m/s}$ at an angle of $30^\circ$ above the horizontal. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the total time of flight of the cannonball before it returns to ground level?

2.0\text{ s}

1.0\text{ s}

4.0\text{ s}

0.5\text{ s}

Show answer & explanation

Answer: $2.0\text{ s}$

Answer

2.0\text{ s}

The correct answer of

2.0\text{ s}

is determined by resolving the initial speed into its vertical component

u_y = 20 \sin 30^\circ = 10\text{ m/s}

and using the total time of flight formula

T = \frac{2 u \sin \theta}{g} = \frac{2(10)}{10} = 2.0\text{ s}

.

Step-by-Step Solution

1

Calculate the vertical component of the initial velocity (

u_y

).

u_y = u \sin \theta = 20 \times \sin 30^\circ = 20 \times 0.5 = 10\text{ m/s}

Only the vertical component of velocity determines the time the projectile remains in the air.

2

Calculate the total time of flight (

T

) using the kinematic formula for full trajectory.

T = \frac{2 u_y}{g} = \frac{2 \times 10}{10} = 2.0\text{ s}

The total time of flight includes both the time to ascend to peak height and descend back to the launch height under gravity

g

.

Key Concept

Total Time of Flight in Projectile Motion

Question 14Question →

A particle is projected from ground level at an angle to the horizontal. At its highest point of trajectory, which of the following statements correctly describes its velocity and acceleration?

The vertical component of velocity is zero, the horizontal component of velocity is non-zero, and the acceleration is directed downward.

Both the total velocity and the acceleration of the particle are zero.

The total velocity of the particle is zero, but its acceleration is directed downward.

The vertical component of velocity is at its maximum value, and the acceleration is zero.

Show answer & explanation

Answer: The vertical component of velocity is zero, the horizontal component of velocity is non-zero, and the acceleration is directed downward.

Answer

At the highest point of projectile motion, the vertical velocity component is zero, the horizontal velocity component is non-zero, and acceleration due to gravity acts downward.

At the peak of a projectile trajectory, vertical upward velocity is fully decelerated by gravity to zero. However, horizontal velocity is maintained because there is no horizontal acceleration. Acceleration due to gravity continues to act constantly downward.

Step-by-Step Solution

1

Analyze vertical velocity at the highest point

v_y = 0

At maximum height, the upward vertical component of motion reduces to zero before the particle begins descending.

2

Analyze horizontal velocity throughout motion

v_x = u \cos\theta \neq 0

In the absence of air resistance, no horizontal force acts on the projectile, keeping horizontal velocity constant throughout flight.

3

Determine the direction of acceleration

a = g

acting vertically downward

Gravity is the only force acting on a projectile in motion, providing continuous downward acceleration.

Key Concept

Velocity components and acceleration at the apex of projectile motion

Estimated Time:45s

Question 15Question →

A shell is launched from level ground into the air. It reaches a maximum height of $45\text{ m}$ above the ground and has a total horizontal range of $240\text{ m}$ . Taking $g = 10\text{ m/s}^2$ , calculate the magnitude of the initial launch velocity of the shell in $\text{m/s}$ .

Show answer & explanation

Answer: 50

Answer

The initial launch velocity of the shell is

50\text{ m/s}

.

Combining the expressions for maximum height

H = \frac{u^2\sin^2\theta}{2g}

and range

R = \frac{2u^2\sin\theta\cos\theta}{g}

gives

\tan\theta = \frac{4H}{R}

. With

H = 45\text{ m}

and

R = 240\text{ m}

, we get

\tan\theta = 0.75 = \frac{3}{4}

, which yields

\sin\theta = 0.6

. Substituting these into the height equation yields

45 = \frac{u^2(0.6)^2}{20}

, solving to

u = 50\text{ m/s}

.

Step-by-Step Solution

1

Express the launch angle in terms of maximum height and horizontal range

\tan\theta = \frac{4H}{R} = \frac{4 \times 45}{240} = 0.75

Dividing the maximum height formula

H = \frac{u^2\sin^2\theta}{2g}

by the horizontal range formula

R = \frac{2u^2\sin\theta\cos\theta}{g}

yields

\frac{H}{R} = \frac{1}{4}\tan\theta

.

2

Find the sine of the launch angle from the tangent value

\sin\theta = 0.6

For a right-angled triangle with

\tan\theta = \frac{3}{4}

, the hypotenuse is

\sqrt{3^2 + 4^2} = 5

, giving

\sin\theta = \frac{3}{5} = 0.6

.

3

Calculate the magnitude of the initial velocity

u

u = 50\text{ m/s}

Substituting values into

H = \frac{u^2\sin^2\theta}{2g}

gives

45 = \frac{u^2 (0.6)^2}{2(10)} \implies 900 = 0.36 u^2 \implies u = 50\text{ m/s}

.

Key Concept

Interdependence of Maximum Height, Range, and Launch Velocity in Projectile Motion

Estimated Time:1m 30s

Question 16Question →

An object is projected from level ground with an initial speed of $30\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the maximum height reached by the object in meters.

Show answer & explanation

Answer: 11.25

Answer

The maximum height reached by the object is

11.25\text{ m}

.

The vertical component of initial velocity is

u_y = 30 \sin(30^\circ) = 15\text{ m/s}

. At maximum height, vertical velocity becomes zero, so

H = \frac{u_y^2}{2g} = \frac{15^2}{20} = 11.25\text{ m}

.

Step-by-Step Solution

1

Calculate the vertical component of the initial velocity.

u_y = u \sin\theta = 30 \times \sin(30^\circ) = 15\text{ m/s}

Only the vertical component of initial velocity determines the maximum height.

2

Apply the vertical motion equation at maximum height where vertical velocity is zero.

H = \frac{u_y^2}{2g} = \frac{15^2}{2 \times 10} = 11.25\text{ m}

Using

v_y^2 = u_y^2 - 2gH

with

v_y = 0

gives

H = \frac{u_y^2}{2g}

.

Key Concept

Maximum height of a projectile

Estimated Time:45s

Question 17Question →

A projectile is launched from level ground with an initial speed of $25\text{ m/s}$ at an angle $\theta$ to the horizontal such that $\sin\theta = 0.80$ . Calculate the maximum height reached by the projectile in meters. [Take $g = 10\text{ m/s}^2$ ]

Show answer & explanation

Answer: 20

Answer

The maximum height reached by the projectile is

20\text{ m}

.

The vertical component of the initial launch velocity is

u_y = u \sin\theta = 25 \times 0.80 = 20\text{ m/s}

. Using the equation for maximum height

H = \frac{u_y^2}{2g}

, we substitute

u_y = 20\text{ m/s}

and

g = 10\text{ m/s}^2

to obtain

H = \frac{400}{20} = 20\text{ m}

.

Step-by-Step Solution

1

Calculate the initial vertical velocity component (

u_y

)

u_y = 25\text{ m/s} \times 0.80 = 20\text{ m/s}

Only the vertical component of velocity determines the maximum height reached.

2

Calculate the maximum height (

H

) using kinematic equations

H = \frac{u_y^2}{2g} = \frac{20^2}{2 \times 10} = 20\text{ m}

At maximum height, the vertical component of velocity becomes zero.

Key Concept

Maximum height of a projectile depends entirely on its initial vertical component of velocity and acceleration due to gravity.

Question 18Question →

A stone is projected horizontally with a speed of $15\text{ m/s}$ from the top of a vertical cliff of height $20\text{ m}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the magnitude of the velocity of the stone just before it strikes the ground?

15\text{ m/s}

20\text{ m/s}

25\text{ m/s}

35\text{ m/s}

Show answer & explanation

Answer: $25\text{ m/s}$

Answer

The magnitude of the velocity of the stone just before hitting the ground is

25\text{ m/s}

.

For a horizontally launched projectile, the horizontal velocity component remains constant at

v_x = 15\text{ m/s}

. The vertical component just before impact is found using

v_y^2 = u_y^2 + 2gh = 0 + 2(10)(20) = 400

, giving

v_y = 20\text{ m/s}

. Combining these perpendicular components yields a total speed of

v = \sqrt{15^2 + 20^2} = 25\text{ m/s}

.

Step-by-Step Solution

1

Identify horizontal velocity component

v_x = 15\text{ m/s}

Air resistance is neglected, so horizontal velocity remains constant throughout flight.

2

Calculate vertical velocity component just before impact using third equation of motion

v_y^2 = u_y^2 + 2gh = 0^2 + 2(10)(20) = 400 \implies v_y = 20\text{ m/s}

Initial vertical velocity

u_y = 0\text{ m/s}

for horizontal projection; stone falls through a vertical displacement of

20\text{ m}

under gravity.

3

Compute total resultant velocity magnitude using Pythagorean theorem

v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ m/s}

Horizontal and vertical velocity components are mutually perpendicular.

Key Concept

Horizontal Projection and Resultant Velocity Vector Synthesis

Estimated Time:2m 0s

Question 19Question →

A projectile is launched from ground level with an initial speed of $40\text{ m/s}$ at an angle $\theta$ to the horizontal such that $\sin\theta = 0.8$ and $\cos\theta = 0.6$ . Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the projectile in $\text{m/s}$ when it reaches a height of $35\text{ m}$ above the ground?

Show answer & explanation

Answer: 30

Answer

The speed of the projectile at a height of

35\text{ m}

above the ground is

30\text{ m/s}

.

The correct answer is

30\text{ m/s}

. The vertical component of velocity at height

35\text{ m}

is found using

v_y^2 = u_y^2 - 2gh = 32^2 - 2(10)(35) = 324

, giving

v_y = 18\text{ m/s}

. Since horizontal velocity component remains constant at

v_x = 24\text{ m/s}

, the overall speed magnitude is given by

v = \sqrt{v_x^2 + v_y^2} = \sqrt{24^2 + 18^2} = \sqrt{900} = 30\text{ m/s}

.

Step-by-Step Solution

1

Resolve initial velocity into horizontal and vertical components

u_x = 24\text{ m/s}

and

u_y = 32\text{ m/s}

Horizontal component

u_x = u\cos\theta = 40 \times 0.6 = 24\text{ m/s}

and vertical component

u_y = u\sin\theta = 40 \times 0.8 = 32\text{ m/s}

.

2

Calculate vertical velocity component at height

h = 35\text{ m}

v_y = 18\text{ m/s}

Using

v_y^2 = u_y^2 - 2gh

, we get

v_y^2 = 32^2 - 2(10)(35) = 1024 - 700 = 324

, yielding

v_y = 18\text{ m/s}

.

3

Calculate the magnitude of total velocity at height

h = 35\text{ m}

v = 30\text{ m/s}

Because air resistance is neglected, horizontal velocity remains constant (

v_x = u_x = 24\text{ m/s}

). Total speed is

v = \sqrt{v_x^2 + v_y^2} = \sqrt{24^2 + 18^2} = 30\text{ m/s}

.

Key Concept

Independence of perpendicular velocity components and calculation of instantaneous speed in projectile motion.

Question 20Question →

A projectile is launched from level ground with an initial velocity of $50\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the projectile at the highest point of its trajectory?

0\text{ m/s}

25\text{ m/s}

25\sqrt{3}\text{ m/s}

50\text{ m/s}

Show answer & explanation

Answer: $25\text{ m/s}$

Answer

The speed of the projectile at its highest point is

25\text{ m/s}

.

In 2D projectile motion, the horizontal component of velocity remains constant throughout flight because no horizontal force acts on the object. At the highest point, vertical velocity reduces to zero, making the total speed equal exclusively to the horizontal velocity component:

v = u \cos 60^\circ = 50 \times 0.5 = 25\text{ m/s}

.

Step-by-Step Solution

1

Resolve the initial velocity into horizontal and vertical components

u_x = u \cos\theta = 50 \cos 60^\circ = 25\text{ m/s}

and

u_y = u \sin\theta = 50 \sin 60^\circ = 25\sqrt{3}\text{ m/s}

Projectile motion decomposes into independent horizontal (constant velocity) and vertical (uniform acceleration) components.

2

Determine the vertical component of velocity at maximum height

v_y = 0\text{ m/s}

At the apex of the parabolic path, the upward vertical motion momentarily stops before descending.

3

Calculate total speed at the apex using vector magnitude

v = \sqrt{v_x^2 + v_y^2} = \sqrt{25^2 + 0^2} = 25\text{ m/s}

Because horizontal acceleration is zero (ignoring air resistance),

v_x = u_x = 25\text{ m/s}

throughout the flight.

Key Concept

Horizontal Component of Velocity in Projectile Motion

Projectile Motion

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept