Question

A tennis ball is dropped on the floor from a height of 20 m . it rebound to a height of 5 m . the ball was in contact with the floor for 0.01 s .what was its average acceleration during contact?​

Answer 1

Answer:

hi u r answer

Explanation:

Favg=(m$\sqrt{2+10+10}$+m$\sqrt{2*10*5}$)/0.01

∴Aavg=$\frac{10(\sqrt{2}+1)}{0.01}$=3000m/s²