Question

A ball of mass 400g is dropped from a height of 5m. A boy on the ground hits the ball vertically upward with a bat with an average force of 100 N so that it attains a vertical height of 20m. The time for which the ball remains in contact with the bat is. (take g=10m/s^2)

Answer 1

m= 0.4 kg     h1 = 5m   g = 10 m/s^2
                     h2 = 20 m
Velocity of the ball after dropping 5m :  u
Velocity of the ball after being hit by the bat:  v
u² = 2 g h1  =>  u = √(2*10*5) = 10 m/s  (downwards)
v² = 2 g h2 = (2 *10 *20)   =>  v = 20 m/s  (upwards)

change of momentum = impulse = m(v+u) = 12 kg-m/s
Force = 100 N = 12 / t
t = time of contact = 0.12 sec

Answer 2

Explanation:

M=0.4gm =0.4kg

h 1=5m , h 2=20m ,g=10m/s

Velocity of the ball after dropping 5m:u

Velocity of the ball after being hit by the bat:v

u 2=2gh M=0.4gm =0.4kg h 1 =5m , h

2

=20m

g=10m/s

2

Velocity of the ball after dropping 5m:u

Velocity of the ball after being hit by the bat:v

u

2

=2gh

1

;u=

(2×10×5)

=10m/s(downwards)

v

2

=2gh

2

;(2×10×20)=v=20m/s(upwards)

Change of momentum =impulse= m(v+u)=12kgm/s

Force=100N=12/t

t= time of contact =0.12sec

Hence,

option (A) is correct answer.

1

;u=

(2×10×5)

=10m/s(downwards)

v

2

=2gh

2

;(2×10×20)=v=20m/s(upwards)

Change of momentum =impulse= m(v+u)=12kgm/s

Force=100N=12/t

t= time of contact =0.12sec