Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
A rocket moves upward, starting from rest with an acceleration of 35.0 m/s2 for 5.63 s. It runs out of fuel at the end of the 5...Question
A rocket moves upward, starting from rest
with an acceleration of 28.1 m/s2
for 5.21 s.
It runs out of fuel at the end of the 5.21 s, but
does not stop.
How high does it rise above the ground?
with an acceleration of 28.1 m/s2
for 5.21 s.
It runs out of fuel at the end of the 5.21 s, but
does not stop.
How high does it rise above the ground?
Answers
bobpursley
Assume full mass of rocket is m, before igniting.
If one assumes the mass of the rocket does not change when burning fuel..
nettrust*time=mass*velocityatburnout
ma*time=m*vburnout
28.1*5.21=vburnout.
how high has it went at burnout?
Vf=2ah where vf=burnout velocity.
solve for h.
Now how much higher does it go?
Vf2^2=Vburnout^2-2gh'
solve for h', add it to h.
vf2=0 at top
If one assumes the mass of the rocket does not change when burning fuel..
nettrust*time=mass*velocityatburnout
ma*time=m*vburnout
28.1*5.21=vburnout.
how high has it went at burnout?
Vf=2ah where vf=burnout velocity.
solve for h.
Now how much higher does it go?
Vf2^2=Vburnout^2-2gh'
solve for h', add it to h.
vf2=0 at top