Question

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

Answer 1

Answer:

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

Explanation:

Using the equations of 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 is 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

Hope this helps!!!