Question

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 6.70 later.

What was the rocket's acceleration?

Answer 1

Is that means bolt's mass is neglected?

Answer 2

I guess so.

Answer 3

If rocket's initial velocity considered as zero:
First, we can find distance traveled by bolt in 6,70 s under g.
y_b=1/2 *g*t*t
Then, that distance is exactly equal to distance traveled by rocket in 4 s.
y_r=y_b=1/2 *a*t*t
substituting t and y_b, we can find a.
Hope this help :)

Answer 4

Thank you. It helps alot. :D

Answer 5

It's a pleasure to help you :)

Answer 6

First, we can find distance traveled by bolt in 6,70 s under g.
y_b=1/2 *g*t*t ?????
how can the initial velocity of bolt be zero?
when it falls off ...it must have some velocity upward and the velocity of bolt will be equal to velocity of rocket at that instant..

Answer 7

@akash123 great! I completely forgot that one.
Let me try this again :)

Answer 8

My apology for previous mistake.
First, let's assume rockets velocity at that instant v, then
v=a*t1 (a is rocket acceleration, t1=4 s) ................. (1)
Distance traveled by rocket in that time:
y=1/2 a t1^2 (t = 4s) ............................................ (2)
When the bolt released, its velocity is equal (1)
so, we can write the height of the bolt as:
y=v*t2-1/2 g*t2^2 (where t2=6,75)...........................(3)
Finally, (2) equals (3) because it's the same height we're talking about:
1/2 a*t1^2=a*t1*t2-1/2 g*t2^2
which you can solve for a :D