Question

In basketball, hang time is the time that both of your feet are off the ground during a jump. The equation for hang time is t = 2 (StartFraction 2 h Over 32 EndFraction) Superscript one-half , where t is the time in seconds, and h is the height of the jump, in feet.

Player 1 had a hang time of 0.9 s. Player 2 had a hang time of 0.8 s. To the nearest inch, how much higher did Player 1 jump than Player 2?

Answer 1

Considering the hang time equation, it is found that Player 1 jumped 0.68 feet higher than Player 2.

What is the hang time equation?

The hang-time of the ball for a player of jump h is given by:

$t = 2\left(\frac{2h}{32}\right)^{\frac{1}{2}}$

The expression can be simplified as:

$t = 2\sqrt{\frac{h}{16}}$

For a player that has a hang time of 0.9s, the jump is found as follows:

$0.9 = 2\sqrt{\frac{h}{16}}$

$\sqrt{\frac{h}{16}} = \frac{0.9}{2}$

$(\sqrt{\frac{h}{16}})^2 = \left(\frac{0.9}{2}\right)^2$

$h = 16\left(\frac{0.9}{2}\right)^2$

h = 3.24 feet.

For a player that has a hang time of 0.8s, the jump is found as follows:

$0.8 = 2\sqrt{\frac{h}{16}}$

$\sqrt{\frac{h}{16}} = \frac{0.8}{2}$

$(\sqrt{\frac{h}{16}})^2 = \left(\frac{0.8}{2}\right)^2$

$h = 16\left(\frac{0.8}{2}\right)^2$

h = 2.56 feet.

The difference is given by:

3.24 - 2.56 = 0.68 feet.

More can be learned about equations at https://brainly.com/question/25537936

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