You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected. A. At what height above your hand does the ball have half as much upward velocity?
B. At what height above your hand does the ball have half as much kinetic energy as when it left your hand?

Explanation:

A ) initial velocity u = 12 m /s

final velocity v = 6 m /s

height = h

acceleration = - g = - 9.8 m /s²

v² = u² - 2gh

6² = 12² - 2 x 9.8 x h

h = 5.51 m

B )

Let the final velocity when energy becomes half be V at height H

kinetic energy at height h = 1/2 m V²

Given ,

.5 x 1 / 2 m x 12² = 1/2 m x V²

V² = 12² / 2

V = 8.486 m /s

V² = u² - 2 gH

8.486² = 12² - 2 x 9.8 x H

H = 3.67 m .  

an energy transformation flow diagram is shown

where is it ?

see earlier