Question

Choose the correct answer:

1. 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 ms–2]

• A. 10
• B. 15
• C. 20
• D. 30

Answer 1

Answer: (D)

30

Answer 2

The correct option is(D): 30 m/s.

Let us assume that they meet at t = t 0

$A:80=\frac{1}{2}gt^2_0....(i)$

$B:80=u(t_0-2)+\frac{1}{2}+\frac{!}{2}g(t_0-2)^2...(ii)$

⇒ 80 = 2 u + 5(2)2

⇒ u = 30 m/s