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Bolt falls off rocket, find rocket acceleration 
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#1
Apr1005, 05:36 PM

P: 13

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 dont understand how to set up this problem, ive never done something like this , asked some friends and they didnt know how to do it. Any help would be greatly appreciated 


#2
Apr1005, 06:13 PM

P: 788

Work backwards kind of. If it took 6.7 for the bolt to hit the ground, how high was it?
[tex]d = v_{i}t + \frac{1}{2}at^2[/tex] 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. [tex]d = v_{i}t + \frac{1}{2}at^2[/tex] d = part a vi = 0 t = 4 Jameson 


#3
Apr1005, 08:10 PM

Mentor
P: 41,316

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: [itex]h = 1/2 a t^2[/itex] and [itex]v = a t[/itex] 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: [itex]h = h_0 + v_0t (g/2)t^2[/itex], using [itex]h_0[/itex] and [itex]v_0[/itex] (in terms of a) from part 1; set h = 0 when t = 6.7 seconds. Solve for a. 


#4
Apr1005, 11:19 PM

P: 13

Bolt falls off rocket, find rocket acceleration
thank you guys



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