Question

A stone is dropped from a height h simultaneously another stone is throw up from the ground which eeaches a height 4h the stone will cross each other after time

Answer 1

Thanks for the question
for the stone going downward the distance traveled will be
 
S1 =ut +1/2gt^2
Here u is 0
G is representing the gravitational acceleration which is equal to 9.8m/s2
So S1= 1/2gt^2
also the distance of stone from the ground level is.
h-S1= h- 0.5 gt^2

Now talk about other stone which is thrown up to height of 4h
Potential energy = kinetic energy

So according to formulas kinetic and potential enrgy
mg(4h) =1/2 mu^2 
u=(8hg)^1/2
So acoording to the law S2= ut - 0.5gt^2

putting value of u here
S2= (8hg)^1/2 t - 0.5t^2
Now in order to these two substances to meet and cross
h -s1=  s2
aslo, h -0.5gt^2 = (8hg)^1/2t - 0.5gt^2   (bu putting values of S1 and S2)

h= (8hg)^1/2t
 t =(h/8g)^1/2     this will be time when stones cross each other

Answer 2

v2 of projection=

v^2-u^2=-2g4h

since v=0

v=sq rt 8gh

time=distance/speed

=h/sq rt8gh

=sq rt h/8g