Two balls are projected simultaneously with the same speed from the top of a tower - one upwards and the other downwards. If they reach the ground in 6s and 2s, the height of the tower is (g = 10 ms^–2)

Lets consider u initial velocity of the ball

Take, g = 10m/s²
From 2nd equation of motion at t = 2sec

s = ut + 1/2​gt²

s = 2u + 1/2​×10×2²

s = 2u + 20. . . . . . .(1)

The time taken to come back at the projection point is 6 − 2 = 4sec

The time to reach the maximum point is 4/2 ​= 2sec

at the maximum height final velocity is zero, v = 0m/s

v = u − gt

0 = u − 2×10

u = 20m/s

From equation 1, we get

s = 2×20 + 20 = 60m