A golf ball is dropped from a height of 1.0 m onto a concrete floor.The velocity of the ball immediately before hitting the floor is 4.43 m/s downward.The ball contacts the floor for 0.940 ms (check out R.Cross,Am.J.Phys.,Vol.67,No.3,March 1999)and leaves the floor traveling upward at 3.96 m/s.The average acceleration of the golf ball while it is in contact with the floor is A)9.81 m/s². B)8.93 * 10³ m/s². C)500 m/s². D)4.71 * 10³ m/s². E)4.21 * 10³ m/s².

Question

Quizplus · Accepted Answer

We can use the formula:
a = (v_f - v_i) / t
where v_i = -4.43 m/s (downward), v_f = 3.96 m/s (upward), and t = 0.940 ms = 0.940 × 10^-3 s (the time the ball is in contact with the floor). 
Note that we need to use the same direction for velocities and acceleration. So we take upward as positive. 
Plugging in the values, we get: 
a = (3.96 - (-4.43)) / 0.940 × 10^-3 
a = 8.93 × 10^3 m/s^2 = 8.93 × 10 m/s^2 = 8.93 m/s^2 (rounded to two decimal places) 
Therefore, the average acceleration of the golf ball while it is in contact with the floor is 8.93 m/s^2, which is closest to choice B.

A Golf Ball Is Dropped from a Height of 1