To test the quality of a tennis ball, you drop it onto the floor from a height of 4.31 m. It rebounds to a height of 2.51 m. If the ball is in contact with the floor for 12.6 ms, what is its average acceleration during that contact?

Answer 1

mgh₁ = mv₁²/2
v₁=sqrt(2gh₁) = sqrt(2•9.8•4.31)=9.2 m/s
mv₂²/2= mgh₂
v₂ =sqrt(2gh₂) = sqrt(2•9.8•2.51)=7 m/s
a=[v(fin) – v(in)]/t
v(fin) =v₂ (directed upward )
v(in)] = v₁ (directed downward )
a= {v₂ - (-v₁)}/t ={v₂ +v₁)}/t=
=(9.2+7)/12.6•10⁻³=1285 m/s² (directed upward)