# A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energy will the ball have at the top of its flight?

Jun 21, 2016

$P E = 43.9 J$

#### Explanation:

At the beginning of the ball's flight, all of its energy can be considered kinetic energy. As the ball moves up, it will slow down as its kinetic energy is converted into potential energy.

At the top of its trajectory, the velocity of the ball is $0 \frac{m}{s}$. This is the point at which ALL of the original kinetic energy has been converted into potential energy. In other words, energy is conserved. Mathematically, this can be written as:

$| \Delta P E | = | \Delta K E |$
$| \Delta P E | = | \frac{1}{2} \cdot m \cdot {v}_{f}^{2} - \frac{1}{2} \cdot m \cdot {v}_{i}^{2} |$
$| \Delta P E | = | 0 J - \frac{1}{2} \cdot 0.19 k g \cdot {\left(21.5 \frac{m}{s}\right)}^{2} |$
$| \Delta P E | = | 0 J - \frac{1}{2} \cdot 0.19 k g \cdot 462.23 {m}^{2} / {s}^{2} |$
$| \Delta P E | = | 0 J - 43.9 J |$
$| \Delta P E | = 43.9 J$