Question

A ball of mass 50g is dropped from a height of 20m.A boy on the ground hits the ball vertically upwardswith a bat with an average force of 200N,so that it attains a vertical height of 45m.The time for which the ball remains in contact with the bat is?(Take g=10m/s²)​

Answer 1

The time for which the ball remains in contact with the bat is?(Take g=10m/s²)

Given:

A ball of mass 50g is dropped from a height of 20m.A boy on the ground hits the ball vertically upwardswith a bat with an average force of 200N,so that it attains a vertical height of 45m.

Before hitting the bat :

Let's take the velocity before hitting it as V1

• V1²
• = 0² + 2gs
• = 2 × 10 × 20
• = 400

• V1 = √400
• V1 = -20 m/s (negative because its downwards)

The speed just after being hit by the bat :

Let's take the velocity after hitting it as V2

Vertical height attained = 45m

• 0
• = V2² - 2gs
• = V2² = 2 × 10 × 45
• = 900
• V2 = 30 m/s (positive because its upward)

Change in Velocity :

• V2 - V1
• = 30 - (-20)
• = 30 + 20
• = 50 m/s

Change in momentum :

• mV2 - mV1 = m∆V
• = 50/1000 × 50
• = 2.5 kg m/s

Now,

• Force = ∆p/∆t = 200 N
• ∆t = 2.5/200
• ∆t = 1/80 sec ans