A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10?3 s). (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Step-by-step solution 58PE Step 1 of 5 Use following equation of kinematics to find the final velocity v of the ball, Rewrite the above equation, Where, is the initial positi n of the ball, is the final positio n of the ball, is the acceleration of the ball and is the initial velocity of the ball. Step 2 of 5 (a) The ball is freely falling under the influence of gravitational force. So, its acceleration is . Substitute, 0 m /s for , for a, 0 m for and 1.50 m for . The velocity is considered as negative because the ball is falling downwards. Hence, the velocity of the ball is . Step 3 of 5 (b) The ball is travelling upwards with retardation due the gravitational force on it. The initial velocity of the ball is calculated as, Rewrite the above equation, Substitute, 0 m/s for v, for a, 1.10 m for and 0 m for . The velocity is considered as positive because the ball is moving upwards. Hence, the velocity of the ball after it strikes the ground is .