 # Ques) Two identical balls are shot upwards one after another at an interval of 2 seconds along the same vertical line with same initial velocity of 40 m/s. Find the height at which the balls collide.

13 years ago

by using the equation of moyion we see that the 1st ball reaches the topmost point i.e 80m above the ground in 4s. again using the equations of motion, we shall find that the moment the first ball reaches its topmost point, the second ball will be at a height of 60m from ground and at that instant its velocity would be 20m/s.

Now that instant should be our t=0 and reference point A. let at a distance of x m above A the balls colide and let the time be t. Then the equation of motion for the second ball will be

x=20t-5t^2.......................(i)

for the motion of the first ball the equation of motion will be

20-x=5t^2........................(ii)

Comparing the above 2 eqs, we have.

20t-5t^2=20-5t^2..............

on solving we have, t=1s.

hence from eqs(i) we have x=15m. so the height will be 75m above the ground.