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= 18s. What is the value of v? ( Take g= 10m/s2)

Answer 1

Answer: 75m/s

Step-by-step explanation:

First find the distance the first ball travels.

1/2*10*18^2 = 1620m

We know that both balls meet after 18s so they each have fallen 1620m when they meet. Second ball falls for 6 seconds less so it falls for 12s

plug this it kinematic equation.

1620 = v*12+1/2*10*12^2

1620 = v*12+720

900 = 12v

v = 75m/s