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 (^^)