3.A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s2 . At an altitude of 0.50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

The maximum altitude reached, measured from the point where the engine shuts off = 0.510 km

The maximum altitude reached measured from the ground = 0.5 + 0.510 = 1.01 km

Explanation:

We will use the equations of motion to analyse this motion.

When the rocket is fired from the ground,

u = initial velocity = 0 m/s (since it was initially at rest)

a = 10 m/s²

The engine cuts off at y = 0.5 km = 500 m

The velocity at that point = V

v² = u² + 2ay

v² = 0² + 2(10)(500) = 10000

v = 100 m/s

The velocity at this point serves as the initial velocity for the next phase of the motion

u = 100 m/s

v = final velocity = 0 m/s (at maximum height, velocity = 0)

y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?

g = acceleration due to gravity = - 9.8 m/s²

v² = u² + 2gy

0 = 100² + 2(-9.8)(y)

- 19.6 y = - 10000

y = 510.2 m = 0.510 km

So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km

truck y has a greater velocity than truck x.

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