Question

From the top of a tower 100 m in height a ball is dropped and at the same instant another ball is projected vertically upwards from the ground so that it just reaches the top of the tower. At what height do the two balls pass one another?

Answer 1

Hey dear @User...

Here is Your answer =)

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#)When the ball is dropped from top, consider the height from top in which the balls meet each other be "x"
#)For the fall dropped from top, u=0m/s, g=9.8 m/s2, h=x, and time taken be "t",
∴x= 0.t + 1/2. (9.8). t2 

or, x=4.9t2 ..................(1)
#)Again, for the ball projected upward, h = (100-x), g=9.8 m/s2, time=t (since both meet at same time), u=25m/s,
∴(100-x) = 25.t - 1/2. (9.8). t2 or, (100-x) = 25t - 4.9t2or, x = 100 - 25t+4.9t2..............(2)
#)now, equating (1) and (2), we get,
=>4.9t2 = 100 - 25t+4.9t2or, 4.9t2 = 100 - 25t+4.9t2 or, 0 = 100-25.t
∴ t = 4 seconds, and, putting t=4 in (1), we get,
x = 4.9t2 or, x = 78.4 meters .
∴ Balls meet each other at the height of 78.4 meters from the top of tower or (100-78.4=)21.6metres above the ground after 4 seconds. ..

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Hope it helped you out =)

Thanks (^^)

Answer 2

Hello bro, Answer is in Attachment..