Question

If the tennis ball were in contact with the ground for 0.0108 s, find the acceleration given to the tennis ball by the ground. Answer in units of m/s2.

Answer 1

Answer:

a=(v-v1)/0.0108.

Explanation:

If the tennis ball is making a contact with the ground for 0.0108 seconds then the ball must be dropped from a certain height, say, h. The ball also must be having certain initial velocity u. While after making contact with the ground the ball must get rebounded to a height say h2 with a final velocity of v2. These velocities can be found with the formulae of √2gx. Where x is the respective height.

Then, the acceleration for the impulse will be given as a=(v-v1)/Δt. Where the Δt is the time of impact so, acceleration is a=(v-v1)/0.0108.