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 ms ^-2]:

Options

• 10

• 15

• 20

• 30

Answer 1

30

Explanation:

As ball A is dropped from tower u = 0 m/s

As the meeting point lies 100 m above ground, displacement of ball will be 80 m.

For ball A

u = 0, S = 80 m, a = +g = +10 m/s² , time = t₁

S = ut + `1/2` at²

80 = `0+1/2 xx 10xxt_1^2`

`160/10 = t_1^2`

⇒ t₁ = 4 s

As ball B is thrown after 2 seconds after release of A. Thus, time available for ball B is 2 seconds to cover a distance of 80 m.

Let speed be 'u' m/s, t₂ = 4 – 2 = 2 s, S = 80 m, a = +g = +10 m/s²

∴ 80 = u × 2 + 1 2 × 10 × (2)²

80 = u + 20

2u = 60

⇒ u = 30 m/s.