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A woman on a bridge 77.0 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 6.50 m more to travel before passing under the bridge. The stone hits the water 3.10 m in front of the raft.

2007-08-26 09:19:45 · 2 answers · asked by slowphysicsstudent 1 in Science & Mathematics Physics

2 answers

First find the fall time:
s = at^2/2, so t = sqrt(2s/a) = sqrt(154/9.81) = 3.962 s
The raft moves 6.5 - 3.1 = 3.4 m in this time, so
v = 3.4 / 3.962 = 0.858 m/s

2007-08-28 05:06:05 · answer #1 · answered by kirchwey 7 · 0 0

incomplete information...
there is nothing mentioned about time so how do u find the speed...
infact the raft may not be moving at all

2007-08-26 09:28:30 · answer #2 · answered by Anonymous · 0 0

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