Question

a stone is allowed to fall from top of the tower 50m high and at the same time another stone is projected vertically upward from the ground with a velocity of 25m/s. calculate when the two stone will meet.

Answer 1

A stone dropped from top will accelerate towards earth at 9.8 m/s^2

distance travelled by the stone is

s1=12at2s1=12at2

For the same stone, the distance travelled from ground is

s=50−s1=50−12at2s=50−s1=50−12at2

For the stone thrown upwards

s=25∗t−12at2s=25∗t−12at2

(minus sign because it is retardation)

When they meet, both distance from ground will be same. Therefore equating both we get

50−12at2=25∗t−12at250−12at2=25∗t−12at2

or

50=25∗t50=25∗t or t=2secondst=2seconds

distance from ground = 50−12∗9.8∗22=30.4m50−12∗9.8∗22=30.4m

distance from top = 50−30.4=19.6m50−30.4=19.6m

Therefore it takes 2 seconds for the stones to cross each other at a distance 19.6m from top of the tower.