A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0s later. What was the rockets acceleration.
The answer i got was 14.715m/s^2 which is wrong.
I thought that the bolts initial velocty should equal the rockets final velocity (when the bolt falls off)
So for the bolt:
V0=(9.81)(6) – Vf=58.86m/s – 0m/s
For the rocket:
Vf=V0 + a(4s)
= 0 + 4a
Subbing one into the other:
58.86m/s=4a
a=14.715m/s^2
Any help would be nice.
Ty
1 Answer

The initial speed of the bolt is not 58.86 m/s.
Let a be the acceleration of the rocket.
During the 4 sec lift off, the rocket has reached a heigth of
h = (1/2)*a*t^2
with t=4,
h = (1/2)*a^16
h = 8*a
Its velocity at 4 sec is
v = t*a
v = 4*a
The initial velocity of the bolt is thus 4*a.
During the 6 sec fall, the bolt has the initial velocity V0=4*a and it drops a total height of h=8*a. From the equation of motion,
h = (1/2)*g*t^2 + V0*t
Subsituting h0=8*a, t=6 and V0=4*a into it,
8*a = (1/2)*g*36 – 4*a*6
Solving fot a,
a = 5.52 m/s^2