From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/s find when and where the two ball will meet
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They will meet instinctively. Here's how.
Lets take distance covered by the stone thrown upwards to meet the other as x. Then the other stone covers a distance of (100-x)m.
Let the ball dropped down be b1. So the ball thrown up is b2.
b1 => d = 100-x
g = 9.8 m/s sq.
u = 0
We know, s= ut + 1/2at sq.
Putting values,
100-x = 4.9t sq. ....(1)
b2 => d = x
g = -9.8 m/s sq.
u = 25 m/s
We know, s= ut + 1/2at sq.
Putting values,
x = 25t -4.9t sq. ....(2)
Adding (1) and (2),
100-x +x = 4.9t sq. + 25t - 4.9t sq.
t = 4 secs.
Puuting t = 4 in (1),
100-x = 4.9t sq.
100-x =19.6
x = 80.4 m
So, they meet at 80.4 m from ground after 4 seconds
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