Question

A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 10.0 and 5.00 s, respectively, with no time interval between them. In Stage 1, the rocket fuel provides an upward acceleration of 15.0 m/s2. In Stage 2, the acceleration is 10.0 m/s2. Neglecting air resistance, calculate the maximum altitude above the surface of Earth of the payload and the time required for it to return to the surface. Assume the acceleration due to gravity is constant.
1) Calculate the maximum altitude. (Express your answer to two significant figures.)
2) Calculate time required to return to the surface (i.e. the total time of flight). (Express your answer to three significant figures.)

Answer 1

Answer:

3.7 km

62.7175453187 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 15\times 10^2\\\Rightarrow s=750\ m$

$v=u+at\\\Rightarrow v=0+15\times 10\\\Rightarrow v=150\ m/s$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=150\times 5+\dfrac{1}{2}\times 10\times 5^2\\\Rightarrow s=875\ m$

$v=u+at\\\Rightarrow v=150+10\times 5\\\Rightarrow v=200\ m/s$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-200^2}{2\times -9.81}\\\Rightarrow s=2038.73598369\ m$

The maximum altitude of the rocket is 750+875+2038.73598369 = 3663.73598369 m = 3.7 km

$s=ut+\frac{1}{2}at^2\\\Rightarrow 3663.73598369=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{3663.73598369\times 2}{9.81}}\\\Rightarrow t=27.3301854818\ s$

Time taken to reach the ground from max height 27.3301854818 seconds

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-200}{-9.81}\\\Rightarrow t=20.3873598369\ s$

Time to reach max height during free fall after stage 2 ends is 20.3873598369 s

Total time of flight is 10+5+20.3873598369+27.3301854818 = 62.7175453187 seconds