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A stone drops from the edge of the roof. It passes the window 2 m high in 0.1s . How far is the roof above the top of the window?
please explain with figure
Height of the window (h) = 2mAcceleration due to gravity g = 9.8 m/s2Time taken (t) = 0.1 sThe velocity of the stone when it reaches the top of the window is to be calculated.Applying the II equation of motion,
h=ut+1/2 at^2
2=0.1u+0.049
0.1u=1.951
u= 19.51m/s
Thus, the velocity 19.51 m/s will be the final velocity of the stone when it is covering the distance between the roof and the window.Initially the stone is at rest i.e., u = 0
v^2-u^2=2gh
(19.51)^2=19.6h
=> h= 19.42m
Distance between the roof and the window =19.42 m.
let the distance between roof and wndow be x metres
then time taken to reach it be t
then x=1/2.g.t^2..........................(1)
for .1 secs more than time t,it covers (x+5) metres
hence x+5=1/2.g.(t+.1)^2
x + 5 = 1/2.g.t^2 + 1/2.g.(.1)^2 + g.t.(.1) { by expanding RHS}
x + 5 = x + 1/20 + t {from eq(1),and putting g=10}
5 - .05 = t
t=4.95 secs
hence the stone takes 4.95 sec to reach the window from the roof
so required distance=x=1/2.g.t^2 (from eq(1) )
x=122.5 metres (ANS)
hope i helped you.
Using second eqn of motion we get:
s=0.5gt2
s=5(0.01)
=0.05m
19.0125 m
i am sorry .
i made a mistake.
for .1 sec. it covers 2 m.
hence x+2=x+1/20+t
t=1.95 sec
hence x=1/2.g.t^2
x=19.01 m(ANS)
0.98M IF U=0
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