Question

A ball is projected vertically upward with an initial velocity of 50 ms–1 at t = 0 s. At t = 2 s, another ball is projected vertically upward with same velocity.
At t = _____s, second ball will meet the first ball. (g = 10 ms–2)

Answer 1

The correct answer is : (6)
At t = 2 s, v 1 = 50 – 2 × 10 = 30 m/s
v 2 = v 2
∴ a rel = g – g = 0
$S = \frac{u² - v²}{2g} = \frac{50² - 30²}{2 × 10} = \frac{1600}{20} = 80m$
$∴ v_{rel} = 50 – 30 = 20 m/s$
$∴Δt = \frac{80 }{ 20} = 4s$
Therefore, Required time t = 2 + 4 = 6 s