A research rocket is launched vertically upwards from rest with a constant acceleration of… — JAMB UTME

A research rocket is launched vertically upwards from rest with a constant acceleration of $5.0\text{ m/s}^2$ . At an altitude of $250\text{ m}$ , its engine suddenly fails and the rocket continues to move vertically upward under gravity alone. Calculate the total time, in seconds, taken by the rocket from launch until it reaches its maximum height. (Take acceleration due to gravity $g = 10\text{ m/s}^2$ )

Answer: 15 s

Answer

The total time taken from launch to reach maximum height is

15\text{ s}

The motion occurs in two phases. In phase 1, accelerating uniformly from rest at

5.0\text{ m/s}^2

over

250\text{ m}

yields a velocity of

50\text{ m/s}

10\text{ s}

. In phase 2, moving upward under gravity alone (

10\text{ m/s}^2

) reduces the velocity from

50\text{ m/s}

to rest (

0\text{ m/s}

) in

5\text{ s}

. Adding the durations of both phases gives

10\text{ s} + 5\text{ s} = 15\text{ s}

Step-by-Step Solution

Calculate the rocket's velocity and elapsed time at the moment of engine failure.

Velocity

v_1 = 50\text{ m/s}

and time

t_1 = 10\text{ s}

The rocket accelerates uniformly from rest at

5.0\text{ m/s}^2

over a distance of

250\text{ m}

Calculate the duration of the unpowered upward motion until vertical velocity becomes zero.

Unpowered flight time

t_2 = 5\text{ s}

After engine failure, the rocket acts as a free projectile moving upward against gravity (

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

) with an initial velocity of

50\text{ m/s}

Sum the time intervals of both stages.

Total time

t_{\text{total}} = 10\text{ s} + 5\text{ s} = 15\text{ s}

The total motion consists of two distinct stages: powered acceleration followed by gravitational deceleration.

Key Concept

Multi-stage vertical motion under constant acceleration followed by free-fall under gravity

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