A projectile is launched from ground level over flat terrain. At time t = 2\text{ s} after… — JAMB UTME

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?

Cevap: 240 m

Cevap

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}

Adım Adım Çözüm

Determine the horizontal component of velocity

v_x = 30\text{ m/s}

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

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

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}

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

Anahtar Kavram

Independence of horizontal and vertical components of projectile motion

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