Question

Height of a tower is 100 m long then the ball is projected up ward and same time the ball throw down find when the ball meet

Answer 1

Each ball is moving in a straight line under uniform acceleration and is thus subject to the equation:

S = ut + 1/2 a t^2

… where S is distance travelled, a is the acceleration towards the ground and t is the journey duration.

We know the sum of their journeys is 100m and their journey durations are equal.

So

100 = (25t -1/2 g t^2) + (0t + 1/2 g t^2)

100 = 25t

So t = 4 seconds

Note: This is is the same time that it would have taken had there been no gravity and the thrown ball covered the entire distance itself. That is intuitive as whatever retarding gravity applies to the thrown ball is equal to the speeding up it applies to the dropped ball.

Note 2 (thanks to comment by Victor Mazmanian): We can assess when the thrown ball reaches its maximum height by solving (v = u+ at) for t when v is 0.

0 = 25 - gt

So t is 25/g … roughly 2.5 seconds

Hence at 4 seconds, when the balls meet the thrown ball is already on its way down after already reached its max height and started descending.

Under the effects of gravity the thrown ball reaches a max height of…

25.25/g - 1/2.g.(25/g)^2 = roughly 31.9 m

but has descending back down to a height of …

25.4 - 1/2.g.4^2 = 100–8g … roughly 21.5m, when the balls meet

And the dropped ball this covers the remaining 1/2.g.4^2 = 78.5m