A ball of mass 50 g is dropped from a height of 20 m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 200 N, so that it attains a vertical height of 45 m. The time for which the ball remains in contact with the bat is [Take g=10 m / s 2 ]

Question

A ball of mass 50 g is dropped from a height of 20 m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 200 N, so that it attains a vertical height of 45 m. The time for which the ball remains in contact with the bat is [Take g=10 m / s 2 ] (A) 1 / 20 text th of a second (B) 1 / 40 text th of a second (C) 1 / 80 text th of a second (D) 1 / 120 text th of a second

Tardigrade Admin · Accepted Answer

Using v2=u2+2 a s v1=√2 g(20)=20 m / s v2=√2 g(45)=30 m / s Impulse =F Δ t=m( vecv2- vecv1) ⇒ 200 t=(50/1000)(20-(-30) t=(5/400)=(1/80) s

Q. A ball of mass $50 g$ is dropped from a height of $20 m$ . A boy on the ground hits the ball vertically upwards with a bat with an average force of $200 N$ , so that it attains a vertical height of $45 m$ . The time for which the ball remains in contact with the bat is [Take $g = 10 m / s^{2}$ ]

$1/2 0^{th}$ of a second

$1/4 0^{th}$ of a second

$1/8 0^{th}$ of a second

$1/12 0^{th}$ of a second

Using $v^{2} = u^{2} + 2 a s$
$v_{1} = 2 g (20) = 20 m / s$
$v_{2} = 2 g (45) = 30 m / s$
Impulse $= F Δ t = m (v_{2} - v_{1})$
$\Rightarrow 200 t = \frac{50}{1000} (20 - (- 30)$
$t = \frac{5}{400} = \frac{1}{80} s$

Q. A ball of mass 50g is dropped from a height of 20m. A boy on the ground hits the ball vertically upwards with 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 ]

1/20th of a second

1/40th of a second

1/80th of a second

1/120th of a second