Question

A rocket moves upward, starting from rest with an acceleration of 28.1 m/s2 for 3.01 s. It runs out of fuel at the end of the 3.01 s, but does not stop. How high does it rise above the ground?

Answer 1

Answer:

491.919561014 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow v=0+28.1\times 3.01\\\Rightarrow v=84.581\ m/s$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 28.1\times 3.01^2\\\Rightarrow s=127.294405\ m$

The distance at which the fuel runs out is 127.294405 m

$v^2-u^2=2gs\\\Rightarrow s=\dfrac{v^2-u^2}{2g}\\\Rightarrow s=\dfrac{0^2-84.581^2}{2\times -9.81}\\\Rightarrow s=364.625156014\ m$

The distance covered after the engine turns off is 364.625156014 m

Total distance above the ground is 364.625156014+127.294405 = 491.919561014 m