Answer to Question #233832 in Physics for Sean

Question

A boy tosses a coin upward with a velocity of 14.7 m/s then catches it on his hand. Find the velocity when the coin returns to the hand. 

Answer 1

Let's take the level of boy's arm above the surface of the Earth as the zero level for potential energy of the coin. Then, at the moment he releases it, coin's total mechanical energy consists only on its kinetic energy (ignoring rotation):

$$E_{tot}^{initial} = \dfrac{mv_{initial}^2}{2}$$

where $m$ is the mass of the coin, and $v_{initial} = 14.7m/s$ is its speed.

According to the energy conservation law, the energy should be the same when he catches it.:

$$E_{tot}^{final}=E_{tot}^{initial}$$

Let's assume, the boy catches the coin at the same level he releases it. Then, again, $E_{tot}^{final}$ consists only of the kinetic energy term, and we have:

$$\dfrac{mv_{final}^2}{2} = \dfrac{mv_{initial}^2}{2}$$

From the last equation:

$$v_{final} = v_{initial} = 14.7m/s$$

Answer. 14.7m/s.