Question

a stone is dropped from a height 300m and at the same time another stone is projected vertically upwards with a velocity 100m and find the height when they meet

Answer 1

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