A stone is dropped from top of tower 300m high at the same time another is projected vertically upward with a velocity of 100m/s from foot of tower find when where two stones meet

Dear student

Height from where the stone is dropped= 300m
g = 9.8 m/s²
velocity of projection of ground projectile = 100 m/s
Velocity of projection of the projectile from some height = 0
Let the particles meet at time t and at height h
Thus, height covered by the projectile from the ground
h = ut - ½ gt²
h = 100t -½ x 9.8 x t²
h = 100t – 4.9t2  … (1)
Distance covered by the projectile thrown down in time t will be, since it falls down from the rest u = 0
300 – h = ½ gt2
300 – (100t – 4.9t2 ) = 4.9t2 …by equation (1)
300 = 100t
t = 3s
Thus particles meet each other at 3s
Height at which particles meet
h = 100t – 4.9t2
h = 100x3 – 4.9 x 9
h = 255.9m

Regards
Arun (askIITians forum expert)