A particle is projected from horizontal ground at an initial angle \theta to the horizonta… — JAMB UTME

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?

A
$40\text{ m}$
$80\text{ m}$ Answer
C
$20\text{ m}$
D
$0\text{ m}$

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}

R = 4H

. Substituting

H = 20\text{ m}

yields

R = 80\text{ m}

Step-by-Step Solution

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.

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.

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

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}

4

yields the exact horizontal range.

Key Concept

Kinetic Energy Conservation and Range-Height Relationship in Projectile Motion

Estimated Time:2m 0s

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