Question

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity u at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of u in ms⁻¹ [ use g=10 m/s² ]

Answer 1

Let us assume that they meet at t = t₀

A : 80 = 1/2 gt₀² ....(i)

B : 80 = u(t₀ − 2) + 1/2 g(t₀ − 2)² ....(ii)

From (i),

t₀ = 4

⇒ 80 = 2u + 5(2)²

⇒ u = 30 m/s