Question

A ball of mass 400 g is dropped from a height of 5 m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 newton so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the bat is [g = 10 m/s²]

पर्याय

• 0.12 s

• 0.08 s

• 0.04 s

• 0.02 s

Answer 1

0.12 s

Explanation:

The speed with which the ball strikes the bat,

v₁ = `sqrt(2gh_1) = sqrt(2 xx 10 xx 5)` = 10 m/s

∴ `vecv_1` = – 10 m/s

The magnitude of the rebound velocity, v₂ = `sqrt(2gh_2)`

= `sqrt(2 xx 10 xx 20)`

⇒ v₂ = 20 m/s

∴ v₂ = 20 m/s

F = `m (dv)/(dt) = (m(vecv_2 - vecv_1))/(dt) = (0.4[20 - (- 10)])/(dt)`

∴ dt = `(0.4(20 + 10))/100`

∴ dt = 0.12 s