Question

A batter hits a pop-up straight up over home plate at an initial velocity of 21 m/s. the ball is caught by the catcher at the same height that it was hit. at what velocity does the ball land in the catcher's mitt? neglect air resistance. (take the positive direction to be upwards.)

Answer 1

The ball rises, converting the initial Kinetic Energy (KE) into Potential Energy (PE), until it comes to rest at a given height, at which point all of the KE has been converted to PE. At that point, it begins to accelerate back downward, until it reaches the catcher's mitt, at the same KE with which it was initially hit. So overall no change in energy as there is no difference in height so the speed should be same but as positive direction is taken to be upwards and ball is now going down the speed will be -21m/s

Answer 2

Our intial velocity is -21 m/s

 

The computation for this problem would be:

vf² = vi² + 2ad 

vf = +/- sqrt[vi² + 2ad] 

since d = 0, 

vf = +/- sqrt[vi²] 

vf = +/- vi 

vf = -vi, from the time when the final velocity must be in the opposite direction of the initial velocity, the answer would be:

vf = - 21 m/s