Question

The best rebounders in basketbal have a vertical leap of about 115.19 cm. What is their initial "launch" speed off the ground?

Answer 1

The initial speed is 4.75 m/s

Explanation:

We can solve this problem by using the law of conservation of energy.

In fact, the total mechanical energy of the rebounder during his motion must be conserved.

When he's jumping from ground level, the mechanical energy is just equal to his kinetic energy:

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

where

m is the mass of the rebounder

v is his initial speed

When he reaches the maximum height, the speed becomes zero, so the kinetic energy is zero and therefore the mechanical energy is just potential energy:

$E=U=mgh$

where

$g=9.8 m/s^2$ is the acceleration of gravity

h = 115.19 cm = 1.1519 m is the maximum height

Since the mechanical energy is conserved, we can equate the two expressions:

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

And so we can find the initial speed:

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.1519)}=4.75 m/s$

Learn more about kinetic energy and potential energy:

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