Question

A ball is dropped from a platforn t=0 s from rest after 6 s another ball is dropped with speed v the balls meet at t=18 s what is the value of v

Answer 1

Answer:

The distance traveled by the first ball in 18s is

 h = 1/2gt^2=21×10×(18)^2 = 1620m

Now to meet second ball has to same distance is (18−6) = 12s

So for second ball,

h = vt+1/2gt^2

1620 = v×12+1/2×10×(12)^2

v = (1620−720)/12 = 75m/s