Let v be the velocity of ball just before it comes to the upper part of the window
Using 2nd equation of motion when the ball passing through window
s=(u×t)+(2a×t2)
now
s=2.20m
t=31sec
a=g=9.8sec2m
2.20=(v×31)+(29.8×31×31)
hence, v=4.95secm
Using 3rd equation of motion from starting point to window
v2−u2=2×a×s
u=0 as ball started from rest
v=4.95secm
a=g=9.8sec2m
so,
4.952=2×9.8×s
s=1.25m
Roof is 1.25m above the window
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