A body is projected from ground level with an initial speed of 40\text{ m/s} at an angle o… — JAMB UTME

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

Answer: 2 s

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

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.

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

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