a rocket takes off vertically from the launchpad with no initial velocity but a
ID: 585882 • Letter: A
Question
a rocket takes off vertically from the launchpad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At 15.4 seconds after blast off, the engine fails completely so the only force on the rocket from then on is the pull of garvity.
a) What is the maximum height the rocket will reach above the launchpad?
b) how fast is the rocket moving at the instant before it crashes onto the launchpad?
c) how long after the engine fails does it take for the rocket to crash onto the launchpad?
Explanation / Answer
part a:
for the first 15.4 seconds:
initial veloicty=0 m/s
acceleration =2.25 m/s^2
then final veloicty at the end of 15.4 seconds=initial veloicty+acceleration*time
=0+2.25*15.4=34.65 m/s
distance covered in this time period=initial veloicty*time+0.5*acceleration*time^2
=0*15.4+0.5*2.25*15.4^2=266.805 m
the maximum height will be achieved, when the velocity becomes 0 and then rocket starts falling back towards earth.
so for the next trip, initial veloicty=34.65 m/s
final veloicty=0
acceleration=-9.8 m/s^2
then using the formula, final veloicty^2-initial velocity^2=2*acceleration*distance
==>0-34.65^2=2*(-9.8)*distance
==>distance=34.65^2/19.6=61.2562 m
time tkaen for this trip=(final speed-initial speed)/acceleration=(0-34.65)/(-9.8)=3.5375 seconds
then total height=266.805+61.2562=328.0612 m
b)from the maximum height, the rocket will start its descent towards launchpad
initial speed=0
distance=328.0612 m
acceleration=9.8 m/s^2
then final veloicty^2=2*9.8*328.0612
==>final veloicty=sqrt(2*9.8*328.0612)=80.1873 m/s
time taken to crash from maximum height=(final speed-initial speed)/acceleration=(80.1873-0)/9.8=8.1824 seconds
it will crash with a speed of 80.1873 m/s
c)hence total time taken after engine failure=3.5375+8.1824=11.7199 seconds
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