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) = 2m
Acceleration due to gravity g = 9.8 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 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