college physics
posted by Drew .
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 7.00s later. What is acceleration of rocket.
I believe if you use x(final)= x(initial +v(initial)deltaT+ 1/2adeltatsquared you get something like x=0+0+1/2(9.8)(4)^2 and that will give distance of the rocket when the bolt fell off. then use the same equation only plug in 240 for x final and plug in 4 for T and solve for a, but when you do that you get 30.01 and that is incorrect for some reason.

The rocket starts from ground, where h=0.
Let
a=upward acceleration of rocket (m/s/s)
t1=4 s
t2=7 s from time when holt falls off
u=velocity of rocket (and bolt) after t1 seconds
= a*t1 m/s
=4a m/s
s1=distance from ground after t1 seconds
=0*t1 + (1/2)a(t1)²
=(1/2)a(t1)²
=8a m
The height h of the bolt from the ground after it falls off
h=u*t2+(1/2)(g)(t2)²
=4a*t2 + (1/2)(9.81)(7²)
=28a + 240.345
Therefore
8a = 28a + 240.345
a=6.676 m/s/s
Check:
u=4a=26.70 m/s
s1=8a=53.4 m
h=u*t2+(1/2)(g)(t2)²
=26.70*7 + (1/2)(9.81)(49)
= 53.4 m from the point bolt dropped off = u OK