A ball is dropped from a high rise platform at t =0 starting from rest. After 6 ~s 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 ? (take g=10 ~ms ⁻² )

Step-by-step solution

. For first ball, u =0 s ₁= 1 2 gt ₁^2= 1 2 g (18)^2 For second ball, initial velocity = v aligned & s ₂= vt ₂+ 1 2 gt ^2 & t ₂=18-6=12 ~s & s ₂= v 12+ 1 2 ~g (12)^2 aligned Here, s ₁= s ₂ aligned 1 2 g(18)^2 & =12 v + 1 2 g(12)^2 v & =74 ~ms ⁻¹ aligned