A ball is dropped from a tower and at the same time another ball is projected vertically
upwards from the ground with a velocity 25 m/s. If they cross each other after 5 seconds at a
point 20 m from the ground, Find height of the tower
(1. Point)
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Answer:
Let the ball dropped down is a and the ball thrown up is b. also the stone thrown upwards cover a distance of x and the other covers a distance of (100 –x).
For ball a
u=0
g=10m/s
2
d=(100−x)
Using the equation
s=ut+
2
1
at
2
100−x=5t
2
.........(1)
For ball b
d=x
g=−10m/s
2
u=25m/s
s=ut+
2
1
at
2
x=25t−5t
2
............(2)
Solving equation (1) and (2)
100=25t
t=4seconds
Put the value of t in equation (1)
x=100−80
x=20m
They will meet at distance of 80 m from the ground after t = 4 seconds
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