Question

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 ?​

Answer 1

Answer:

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