Question

A ball is dropped from the 100m height tower simultaneously another ball was thrown upward from the bottom of the yower with a speed 50m/s . they will cross each other

Answer 1

They meet when their distances travelled equal to 100m.

Therefore,

$S_1+S_2=100m$

$(ut+\frac{1}{2} gt^2) +(ut-\frac{1}{2}gt^2)=100m$

As, g will be positive for the falling body and negative for rising body.

u for the falling body is 0, and for the rising body is 50m/s

Therefore,

$5t^2+(50t-5t^2)=100m = 50t = 100m, Or, t = 2s$

Therefore, they cross each other at t = 2s.