A ball is dropped from a height of 7.2 M it bounces back to 3.2 after striking the floor the ball remains in contact with the floor for 20 MS given that G is equal to 10 M per second square the average acceleration of the ball during contact is?

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Answer 1 · 2018-03-24T07:09:09

The acceleration is #=1024.3ms^-2#

If a ball is dropped ftom a height #=h#, the velocity on reaching the ground is given by the equation

#mgh=1/2mu^2#

#u^2=2gh#

#u=sqrt(2gh)#

The acceleration due to gravity is #g=10ms^-2#

The initial velocity on reaching the ground is

#u=sqrt(2*10*7.2)ms^-1#

The velocity on leaving the ground is

#v=sqrt(2*10*3.6)ms^-1#

The change in momentum is

#=m u-(-mv)=m(u+v)kgms^-1#

But the change in momentum is equsal to the impulse

#Ft=Delta(mu)#

#t=20*10^-3s#

#F*20*10^-3=m(sqrt144+sqrt72)#

#F=(m(sqrt144+sqrt72))/(20*10^-3) N#

According to Newton's Secnod Law of motion

#F=ma#

Therefore,

#ma=(m(sqrt144+sqrt72))/(20*10^-3)#

The acceleration is

#a=((sqrt144+sqrt72))/(20*10^-3)#

#=1024.3ms^-2#