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

Question #4144

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.

Expert's answer

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:

h

1.5 = 9.8t + 9.8t

4.9t

t = (-9.8+ 11.2)/9.8 = 0.143

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:

h

_{w }= vt + gt^{2}/21.5 = 9.8t + 9.8t

^{2}/24.9t

^{2}+ 9.8t - 1.5 = 0t = (-9.8+ 11.2)/9.8 = 0.143

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