A body is dropped from a point 4.9m above a window 1.5m high. Find the time taken by the body to pass against the window.
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Expert's answer
2011-09-07T05:21:32-0400
The falling time before passing the window is t = √ 2h1/g = √2*4.9/9.8 = √1 =1 s. The velocity after 1 second falling is v = gt = 9.8 *1 = 9.8 m/s.
The equation of the motion while passing the window is the following: hw = vt + gt2/2 1.5 = 9.8t + 9.8t2/2 4.9t2 + 9.8t - 1.5 = 0 t = (-9.8+ 11.2)/9.8 = 0.143
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