Stone drops from the edge of the roof. It passes window 2m high in 0.1 sec. How far is the roof above the top of the window? (takeg=10m/s2)
Height of the window (h) = 2m
Acceleration due to gravity g=10 m/s2
Time taken (t) = 0.1 s
The velocity of the stone when it reaches the top of the window is to be calculated.
Applying the second equation of motion,
h=ut+12at2
2=u(0.1)+5(0.01)
(0.1)u=1.95
U=19.5m/s
Thus, the velocity 19.5 m/s will be the final velocity as soon as the stone starts covering the distance between the roof and the upper edge of the window.
As stone starts from rest i.e., u = 0
v2−u2=2gh
(19.5)2−0=20h
⇒h=19.01m
Distance between the roof and upper edge of the window = 19.01 m.