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A small ball rolls horizontally off the edge of a tabletop
that is 1.20m high. It strikes the floor at a point 1.52m horizontally from the table edge. (a) How long is the ball in the
air? (b) What is its speed at the instant it leaves the table? Show and explain your work. Thanks :)

2007-09-19 08:47:36 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

You should consider this ball a free falling object when it leaves the edge of a tabletop, since its movement initially is only horizontal. Thus:
0.5*g*t^2 = 1.20
t = sqrt(2.4/9.8) = 0.495(s)
(a) The ball in the air for only 0.495s.
(b) At the instant the ball leaves the table, it only has horizontal speed. Since it strikes the floor at a point 1.52m horizontally from the table edge after 0.495s, its horizontal speed must be:
1.52m/0.495s = 3.07m/s.

2007-09-20 08:40:21 · answer #1 · answered by Hahaha 7 · 1 0

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