Question

A stone is dropped from a height 300m and at the same time another stone projected vertically upward with a velocity of 100m/s . Find when and where the stone 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

Answer 2

This problem can be solved by using Relative motion concept which make it much easier to solve.
For it , Suppose the snapshot when the Stone dropped from the height (say Stone A) and the stone projected from the ground (say Stone B) .
for the stone A :
initial velocity ( u) = 0.
acceleration = g ( downward )
(g = 10 m/s² , acceleration due to gravity)
for Stone B :
initial velocity = 100m/s. ( upwards)
acceleration = g (downward)
Further ,
Let's make one of the stone to be at REST ( but actually it is not at rest ) to apply relative motion concept.
suppose stone A is at rest .
Now new Values of velocity and acceleration for stone B can be determined as follows :
As u can see that both stones are approaching each other .
then,

New initial velocity of stone B(say v) = [100(upwards)-u(downward)],
as u = 0;
v= 100m/s.
New Acceleration for the stone B (say a) = g (downward ) - g (downward)
= g- g = 0.
Further ,
distance between the two stones = 300m,
then,
time taken for stone B to reach stone A is given by : 300/100 = 3 seconds .
Note : in our relative frame the motion is non accelerated .
This is the time when the two stones meet .
#Distance from the ground ( say s) when they meet :
(Trace the motion of Stone B )
using s = ut +½×at²,
s= 100×3 - ½ × 10 × 3²,
s = 255 m .