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# A ball of mass 50g is  dropped from a height of 20m. A boy on the ground hits the ball 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 (g=10m/s^2)A)1/20th ofa second B)1/40th of a secondC)1/80th of a secondD)1/120th of a second

Kelvin Piety DeCosta
29 Points
7 years ago
Velocity of the ball just before getting hit = sqrt(2(-10m/s^2)*(-20m)) = 20m/s
After the ball is hit it reaches a hit 45 m with -10m/s^2 acceleration. Note: The bat force is not counted here as it is temporary.
Velocity just after the ball is struck = sqrt(2*10*45)=30m/s = -30m/s (because it is opposite to earlier reference)
Impulse = change in momentum = 0.05kg(-30-20m/s) = -2.5Ns
F=-200N
t=J/F = 2.5/200 = 1/80
But wait for an expert review of the question!
shubhee
11 Points
4 years ago

From up to down

u1= velocity up = 0

V1= velocity before hitting bat

s1= distance before hitting bat

v1^2 = u1^2 + 2as

v1^2 = 0+ 2*10*45

V1= 20 m/s

F=M*A

A= 200/ 50*10^-3

=4000

U2 = velocity after hitting bat

V2= velocity at top = 0

s2 = distance covered after hitting bat = 45

u2^2= -2*10*45

u2= -30

time :

-30= 20 - 4000*t

t= (-30–20/-4000)

= 1/80th of a second