Question

A rocket takes off vertically from the launchpad with no initial velocity but a constant upward acceleration of 2.5 m/s^2. At 10 s after takeoff, the engines fail completely so the only force on the rocket from then on is the pull of gravity.

A) What is the maximum height the rocket will reach above the launchpad?

B) How long after engine failure does it take for the rocket to crash back down?

Answer 1

A)
To find out how far the rocket will go, first find how high the rocket goes with help from the engine, then find out how high the rocket goes after the engine is cut off.
So:
when engine is on:
a= 2.25m/s^2
vi = 0m/s
t = 10s
d = vit + 1/2at^2
d = 0 + 1/2x2.25x10^2
d= 112.5m

find final velocity, so that you can answer the next part (final velocity of rocket while engine is on becomes initial velocity when engine is off)
vf^2 = vi^2 + 2ad
vf^2 = 0 + 2*2.25*112.5
vf= 22.5m/s

when engine is off, but rocket is still going up:
a = -9.81m/s^2 (force of gravity)
vi = 22.5m/s (vf from before becomes vi)
vf = 0m/s (when the rocket reaches max height, the final velocity will be 0)
to find d, use formula vf^2= vi^2 + 2ad
0 = 22.5^2 + 2*(-9.81)d
solve the equation to get d= 25.8028m
so max height would be 112.5 + 25.8028 = 138.3 m

B)
You already know the time duration while the rocket's engines were working, but now we need to find the time it takes while the rocket is still moving upwards with the engines off, and find how long does it take for the rocket to get back down.
To find how long it takes for the rocket to move upwards while engine is off:
d= 25.8028m
vi=22.5m/s
a= -9.81m/s^2
vf=0 (when the rocket reaches the top, vf is 0)
vf= vi + at
0=22.5 - 9.81t
Solve to get t= 2.29 s
To find the time it takes for the rocket to get from the top to the bottom:
d= -138.3m (top to bottom)
vi=0m/s
a=-9.81m/s^2
d= vit + 1/2at^2
-138.3= 0+ 1/2*(-9.81)t^2
solve for t to get t = 5.31s
so add up all the times to get the total time the rocket was in the air: 10 + 2.29 + 5.31 = 17.6 s