Question

During launches, rockets often discard unneeded parts. A certainrocket starts from rest on the launch pad and accelerates upward ata steady 3.25 m/s^2. When it is 230 m above the launch pad, it discards a usedfuel canister by simply disconnecting it. Once it is disconnected,the only force acting on the canister is gravity (air resistancecan be ignored).
A) How high is the rocket when thecanister hits the launch pad, assuming that the rocket does notchange its acceleration?
B) What total distance did the canistertravel between its release and its crash onto the launchpad?

Answer 1

Answer:

915.69549 m

306.1968 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 3.25 m/s²

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 3.25\times 230+0^2}\\\Rightarrow v=38.665\ m/s$

Height is given by

$h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{38.665^2}{2\times 9.81}\\\Rightarrow h=76.1968\ m$

Total distance the canister has to travel is 230+76.1968 = 306.1968 m

Time to reach the max height

$v=u+at\\\Rightarrow t=\dfrac{v-u}{g}\\\Rightarrow t=\dfrac{0-38.665}{-9.81}\\\Rightarrow t=3.9413\ s$

Time taken to reach the ground is

$s=ut+\frac{1}{2}gt^2\\\Rightarrow 306.1968=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{306.1968\times 2}{9.81}}\\\Rightarrow t=7.9\ s$

Total time taken would be 3.9413+7.9 = 11.8413 seconds

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=38.665\times 11.8413+\dfrac{1}{2}\times 3.25\times 11.8413^2\\\Rightarrow s=685.69549\ m$

The rocket is 685.69549+230 = 915.69549 m from the launchpad.

The total distance is 230+76.1968 = 306.1968 m