Question

A ball of mass 400 gm is dropped from a height of 5m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 N 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)

• A. 0.12 sec
• B. 0.08 sec
• C. 0.04 sec
• D. 12 sec

Answer 1

Answer: (A)

0.12 sec

Answer 2

Velocity of ball just before collision,

v₁ = $\sqrt{2gh_{1}}$ = $\sqrt{2 \times 10 \times 5}$ = 10 m/s

Velocity of ball just after collision,

v₂ = $\sqrt{2gh_{2}}$ = $\sqrt{2 \times 10 \times 20}$ = 20 m/s

 p₁ = mv₁ and p₂ = –mv₂

 p = p₂ – p₁ = –m(v₁ + v₂)

F_av = $\frac{\Delta p}{\Delta t}$ = $\frac{m(v_{1} + v_{2})}{\Delta t}$

 100 = $\frac{0.4(10 + 20)}{\Delta t}$

 t = 0.12 sec