Question

rocket has initial velocity of 35m/s-1.acceleration is 5.0m/s-2.engine break at hight 20km.what the maximum height achived by the rocket

Answer 1

Rocket has initial velocity of $35\mathrm{ms}^{-1}$. Acceleration is $5.0\mathrm{ms}^{-2}$. Engine break at high 20km. What the maximum height achieved by the rocket?

**Solution.**

$$v_{i} = 35 \frac{m}{s}; a = 5.0 \frac{m}{s^{2}}; h_{1} = 20\mathrm{km} = 20 \cdot 10^{3}\mathrm{m}, g = 9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}};$$ $$h_{\max} - ?$$

The height $h_1$ achieved by the rocket with the engine running:

$$h_{1} = \frac{v_{1}^{2} - v_{i}^{2}}{2a};$$

$v_{i}$ - the initial velocity of the rocket;

$v_{1}$ - the final velocity of the rocket when it moved with the engine running;

$a$ - the acceleration of the rocket.

$$v_{1}^{2} = 2a h_{1} + v_{i}^{2}.$$

The height $h_2$ achieved by the rocket without the engine running:

$$h_{2} = \frac{v_{2}^{2} - v_{1}^{2}}{-2g};$$

$v_{1}$ - the initial velocity of the rocket when the engine break;

$v_{2} = 0$ - the final velocity of the rocket when it achieved the maximum high;

$g$ - the gravity acceleration.

$$h_{2} = \frac{0 - v_{1}^{2}}{-2g} = \frac{-v_{1}^{2}}{-2g} = \frac{v_{1}^{2}}{2g};$$ $$h_{2} = \frac{2a h_{1} + v_{i}^{2}}{2g}.$$

The maximum height achieved by the rocket:

$$h_{\max} = h_{1} + h_{2};$$ $$h_{\max} = h_{1} + \frac{2a h_{1} + v_{i}^{2}}{2g}.$$ $$h_{\max} = 20 \cdot 10^{3}\mathrm{m} + \frac{2 \cdot 5.0 \frac{\mathrm{m}}{\mathrm{s}^{2}} \cdot 20 \cdot 10^{3}\mathrm{m} + \left(35 \frac{\mathrm{m}}{\mathrm{s}}\right)^{2}}{2 \cdot 9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}} = 30267\mathrm{m}.$$

**Answer:** The maximum height achieved by the rocket is $h_{\max} = 30267\mathrm{m}$.