Bolt falls off rocket, find rocket acceleration

AI Thread Summary
A rocket is launched with constant acceleration, and a bolt falls off four seconds after liftoff, hitting the ground 6.7 seconds later. To find the rocket's acceleration, the height the bolt falls from must first be calculated using the equation d = v_{i}t + 1/2at^2, with initial velocity set to zero. After determining the height and speed of the rocket at four seconds, these values are used to establish the bolt's motion after it detaches. The final equation for the bolt's height incorporates the rocket's initial height and speed, allowing for the calculation of the rocket's acceleration. This methodical approach leads to the solution for the rocket's acceleration.
stangeroo
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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 s later.

Find the rockets acceleration.

I don't understand how to set up this problem, I've never done something like this , asked some friends and they didnt know how to do it. Any help would be greatly appreciated :smile:
 
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Work backwards kind of. If it took 6.7 for the bolt to hit the ground, how high was it?

d = v_{i}t + \frac{1}{2}at^2

vi = 0
a = -32 ft\sec^2 or -9.8 m\sec^2
t = 6.7 s

Solve for d.

Now, use the same equation and solve for "a" of the rocket.

d = v_{i}t + \frac{1}{2}at^2

d = part a
vi = 0
t = 4


Jameson
 
Do it in two parts, but like this:
Part 1: Find the height and speed of the rocket after 4 seconds in terms of the rocket acceleration. Use:
h = 1/2 a t^2 and
v = a t

Part 2: Realize that the initial height and speed of the bolt equals the height and speed of the rocket after 4 seconds (which was calculated in part 1). The height of the bolt after it leaves the rocket is:
h = h_0 + v_0t -(g/2)t^2,
using h_0 and v_0 (in terms of a) from part 1; set h = 0 when t = 6.7 seconds. Solve for a.
 
thank you guys:cool:
 
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