A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s. What is the value of v?

73.5 m/s

Explanation:

The position of the first ball is:

y = y₀ + v₀ t + ½ at²

y = h + (0) (18) + ½ (-9.8) (18) ²

y = h - 1587.6

The position of the second ball is:

y = y₀ + v₀ t + ½ at²

y = h + (-v) (18-6) + ½ (-9.8) (18-6) ²

y = h - 12v - 705.6

Setting the positions equal:

h - 1587.6 = h - 12v - 705.6

-1587.6 = - 12v - 705.6

1587.6 = 12v + 705.6

882 = 12v

v = 73.5

The second ball is thrown downwards with a speed of 73.5 m/s