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A 0.4 kg tennis ball is dropped from rest at a height of 3.9 m onto a hard floor.

a) What is the speed of the ball at the instant of contact with the floor? I solved to be 8.743

A flash photograph shows that the ball is compressed a maximum of 0.6 cm when it strikes the floor.

b) Assuming that the acceleration of the ball is constant during its contact with the floor, what force does the floor exert on the ball?

I used my v final from part a as my v initial for part b. So

v initial = 8.743

v final = 0

delta x = 3.9

using the formula v

_{f}

^{2}=v

_{I}

^{2}+2a

_{x}delta x

which gave me 0=(8.734)

^{2}+(2)a

_{x}(.6) = 63.7

then using F=ma+9.8m

gave f= (0.4)(63.7)+(9.8)(0.4) so F=29.4

but this is incorrect. Can anyone point out what I'm missing?