Question

Lebron James ( 105 Kg ) runs at a speed of 5.4 m/s before dunking.  How high did he jump?

Answer 1

Given data:

* The mass of the Lebron James is m = 105 kg.

* The value of the speed given is v = 5.4 m/s.

Solution:

The kinetic energy of the Lebron James before the jump is,

$K=\frac{1}{2}mv^2$

The potential energy at the top of the jump is,

$U=\text{mgh}$

where g is the acceleration due to gravity,

According to the law of conservation of energy,

$\begin{gathered} K=U \\ \frac{1}{2}mv^2=\text{mgh} \\ \frac{1}{2}v^2=gh \\ h=\frac{v^2}{2g}^{} \end{gathered}$

Substituting the known values,

$\begin{gathered} h=\frac{5.4^2}{2\times9.8} \\ h=1.5\text{ m} \end{gathered}$

Thus, the height of the jump is 1.5 meters.