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A 500g model rocket is resting horizontally at the top edge of a 40m high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5m/s and at the same time causes the engine to ignite. When the engine fires, it exerts a constant 20N horizontal thrust away from the wall.

a) How far from the base of the wall does the rocket land?
b) Describe the trajectory of the rocket while it travels to the ground.

2007-06-05 04:34:52 · 1 answers · asked by Subaki 1 in Science & Mathematics Physics

1 answers

You need to remember 3 things:
F = ma
s = at²/2 (with no initial velocity)
s = v0T + at²/2 (with initial velocity v0)
When it galls off of the wall (with no initial velocity 'down')
s = at²/2 => t = √(2s/a) = √(2*40/9.8) = 2.857 sec.
The horizontal acceleration is given by
a = F/m = 20/.5 = 40 m/sec² so that, after 2.857 sec. it has travelled
s = .5*2.857 + 40*2.857²/2 = 167.677 meters from the wall. Since it is undergoing constant acceleration in both y --and-- x, it's path will be a straight line (at an angle of about
arctan(40/167.677) = 13.417° below horizontal)

Doug

2007-06-05 05:00:37 · answer #1 · answered by doug_donaghue 7 · 0 0

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