Answer to Question #131332 in Physics for Caleb

Question

A rocket at rat is launched to move upwards. At first 30sec the rocket moves with an upward acceleration of 18ms^-2 then the rocket engine shuts down as a result the rocket keeps on moving upward for a while and then falls back to ground. Calculate the maximum height that the rocket can reach

Answer 1

After the 30 sec of acceleration, the upward velocity of the rocket wil be:

$$v = at = 18m/s^2\cdot 30s = 540 m/s$$

The position above Earth will be:

$$h_1 = \dfrac{at^2}{2} = \dfrac{18\cdot 30^2}{2} = 8100 m$$

Then, the rocket will be moving under the acceleration $g = 9.81 m/s^2$, which is directed downward. The time it takes to stop is:

$$t = v/g = 540/9.81\approx 55.05s$$

During this time the rocket will cover:

$$h_2 = vt - \dfrac{gt^2}{2} = 540\cdot 55.05 - \dfrac{9.81 \cdot 55.05^2}{2} \approx 14862m$$

The total height is

$$h = h_1 + h_2 = 22962m = 22.962 km$$

Answer. 22.692 km.