Question

1. A basketball player can jump 0.72m off the floor from a standing position. Calculate the player’s speed when leaving the floor

Answer 1

The potential energy of the basketball player at the highest point is:

$$U = mgh$$

where $m$ is the mass of the player, $h = 0.72m$ is the height and $g = 9.81m/s^2$ is the gravitational acceleration. Using the energy conservation law find, that this potential energy is equal to the kinetic energy when player was leaving the floor:

$$U = K=\dfrac{mv^2}{2}$$

where $v$ is the player’s speed when leaving the floor. Substituting the expression for $U$ and expressing $v$, obtain:

$$mgh=\dfrac{mv^2}{2}\\ v = \sqrt{2gh}\\ v = \sqrt{2\cdot 9.81\cdot 0.72} \approx 3.76m/s$$

Answer. 3.76 m/s.