From the top of the tower 100m in height, a ball is dropped and at the same time, another ball is projected vertically upward from the ground with a velocity of 25m/s. Find when and where the 2 balls meet(a=9.8m/s2

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.