Free fall acceleration of rocket

[rich1116]

can someone please help me??

I thought I had the answer to this problem, but it doesn't match the book's answer. I can't figure out what I'm doing wrong. This is the problem:

A rocket is launched straight up with constant acceleration. 4 seconds after liftoff, a bolt falls off the rocket and hits the ground 6 seconds later. What was the rocket's acceleration?

First, I used the free fall acceleration of the bolt, -9.80m/s2 and 6 seconds that if fell for to find its position when it fell off the rocket, which I found to be 176.4m. This would also be the height of the rocket when the bolt fell of. Since the rocket traveled for 4 sec to reach this height, I used this height and time to find the acceleration of the rocket, 22.05m/s2.
The book answer however is 5.5 m/s2.
What am I doing wrong?

 

[Dadface]

Hello rich.You took the initial velocity of the bolt to be zero.It is not zero it is the upward velocity that the rocket has at the instant the bolt drops.

 

[rich1116]

then how do i find the initial velocity?

 

[Dadface]

You can write it in symbol form.

 

[rich1116]

i can but symbols won't due me any good. i need a definite numerical answer to the question. i know i can write a bunch of equations but i have no numbers to put into them

 

[Dadface]

Consider the rocket having risen for 4 seconds
u=0, t=4, a=a(unknown)
you can write v=4a and s=8a(equations of motion)
Now consider the bolt
u=4a, a=9.8,t=6, s=8a(this is the displacement not the distance traveled because the bolt rises a bit before it comes to momentary rest and then falls)
Using the values above write down the equation connecting s,u,a(9.8 and not to be confused with a for the rocket)) and t,plug in the numbers and solve.Make sure you get the minus sign in front of ut ie the displacement and acceleration are down whereas u is up.Good luck with it.