A stone is allowed to fall from the top of a tower 100 m
and at the same time another stone is projected vertina
upwards from the ground with a velocity of 25 m/s. Calculat
when and where the two stones will meet.
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Answer:
First check diagram and Follow the below Soln.
Explanation:
Given
In phase B, u=2m/s
AB=100m
Now same time would be reqd. for two balls to meet. Let time be t
Ball in phase B
s=ut
s=25*t =25t m........ (i)
Ball in phase A
s=ut+1/2gt2
u=0
g=10m/s2
s=1/2*10t2
s=5t2 m......... (ii)
From (i) and (ii) we have,
5t2+25t=100
t2+5t=20
t2+5t-20=0
t=2.62s
Reqd. distance =25*2.62=65.5
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