Movement/change in velocity problem?

Question

A student is performing an experiment where she allows a golfball to fall from a height of 2m onto a concrete floor. The student measures that the ball rebounds to a height of 1.6m . An electronic timer gives the contact time with the floor as 0.05s.

Calculate the balls change in velocity as it strikes the floor and rebounds.

I've already found that the speed at which the ball reaches the floor is 6.3 m/s, but i really have no idea what to do from here. The answer in the textbook is 12 m/s up.

Any help is appreciated. Thanks. :)

Answer 1

The velocity with which body strikes the ground is v1 = - 6.3 m/s( taking downward axis as - ve)



As the body rebounds back-

Let u = v2

v = 0

a = - g = - 9.8 m/s2

s = 1.6 m

Using 3rd eqn of motion

v2-u2 = 2as

0-(v2)2 = 2 x -9.8 x 1.6

v2 = 5.6m/s

Change in velocity = v2 -v1 =5.6 - (- 6.3) = 11.9 m/s

Answer 2

let the velocity at instant of release from height of 2m be u = 0m/s

thus velocity when golf ball makes contact with floor is by

  v^2 = u^2 + 2*9.81*h

       = 0 + 2*9.81*2

  v    = 6.2641m/s

similarly,

to find out velocity of ball when it departs from floor use same fomula.

with g negative as motion will be against gravity.

thus velocity of ball when it reaches 1.6m v = 0m/s

    0 = u^2 - 2*9.81*1.6

    u =  5.60m/s

thus change in velocity = initial velocity(at arrival) - fianl velocity(at departure) (note that here consider any one direction positive and thus other negative)

i have assumed downward positive

thus change in velocity = 6.264-(-5.6) = 11.864m/s