Answer to Question #4144 in Mechanics | Relativity for akshaygopal
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.
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