Question

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 m/s2 upward. At 25.0 s after launch, the second stage fires for 10.0 s, which boosts the rocket�s velocity to 132.5 m/s upward at 35.0 s after launch. This firing uses up all the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Air resistance can be neglected.Part AFind the maximum height that the stage-two rocket reaches above the launch pad.?

Answer 1

Answer:

3091.56

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 25+\frac{1}{2}\times 3.5\times 25^2\\\Rightarrow s=1093.75\ m$

Distance traveled in the first stage is 1093.75 m

$v=u+at\\\Rightarrow v=0+3.5\times 25\\\Rightarrow v=87.5\ m/s$

Velocity at the end of first stage is 87.5 m/s

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{132.5-87.5}{10}\\\Rightarrow a=4.5\ m/s^2$

Acceleration of the second stage is 4.5 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=87.5\times 10+\frac{1}{2}\times 4.5\times 10^2\\\Rightarrow s=1100\ m$

Distance traveled in the second stage is 1100 m

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-132.5^2}{2\times -9.81}\\\Rightarrow s=894.81\ m$

Distance traveled after the second stage has stopped firing is 894.81 m

Total height the rocket reached = 1093.75+1100+897.81 = 3091.56 m