Question

a ball of mass 50g is dropped from. a height 20m​

Answer 1

hey there is answer !!!

I hope you help .....

Velocityoftheballjustbeforehittingthebat=V

1

V

1

2

=0

2

+2gs=2×10×20=400

V

1

=−20m/s,V

1

isdownwards

heightattainedbytheballafterbeinghit:−45m

Thespeedjustafterbeinghitbythebat=V

2

⇒0=V

2

2

−2gs

⇒V

2

2

=2×10×45=900

V

2

=30m/sec(upwards)

Changeinvelocity=V

2

−V

1

=30−(−20)=50m/s

Changeinmomentum=mV

2

−mV

1

=mΔV=

1000

50

×50=2.5kgm/s

force=

Δt

Δp

=200N

Δt=

200

2.5

=

80

1

sec

Hence,

option (C) is correct answer.

regards you ........

Answer 2

Answer:

velocity of the ball just before hitting the bat = v1

    v1² = 0² + 2 g S = 2 * 10 * 20 = 400

  v1 = - 20 m/s    ,  v1 is downwards.

Height attained by the ball after being hit :  45 m

The speed just after being hit by the bat.  =  v2 =>

        0 = v2² - 2 g S       =>  v2² = 2 * 10 * 45 = 900

    v2 = 30 m/sec    upwards

Change in velocity = v2 - v1 = 30 - (-20) = 50 m/s

change in momentum = m v2 - m v1 = m Δv = 50 /1000 * 50 = 2.5 kg-m/sec

    force = Δp / Δt  = 200 Newtons

        Δt = 2.5 /200 = 1/80 sec  = 0.0125  sec

hence ,c is correct option

MARK ME BRAINLIEST AND THANKS

Explanation: