Question

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. At what height above your hand does the ball have half as much upward velocity?

Answer 1

Answer:

$h = 5.505\,m$

Explanation:

Let assume that ball begins its movement at a height of zero. The height can be found by energy approach, to be precise, by Principle of Energy Conservation:

$\frac{1}{2}\cdot (0.145\,kg)\cdot (12\,\frac{m}{s} )^{2}= \frac{1}{2}\cdot (0.145\,kg)\cdot (6\,\frac{m}{s} )^{2}+(0.145\,kg)\cdot (9,81\,\frac{m}{s^{2}} )\cdot h$

$h = 5.505\,m$