Question

Formula to find the distance at which two balls collide which are thrown simuntaneously

Answer 1

Suppose they meet at a height h from ground in time t. Their initial velocity of projection be u = 20 m/s

For the ball thrown upwards, (u and g are opposite)

h = ut - gt2 /2

For the ball thrown vertically downwards, (u and g are in the same direction)

40 - h = ut + gt2 /2

Add the two equations

40 = 2ut

t = 40/2u = 40/2x20 = 1 sec

h = 20 x 1 - 9.8 x 12 /2 = 20 - 4.9 = 15.1 m

They meet at height 15.1 m from the ground in 1 second.