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Sagot :
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:
[tex]t = 2\left(\frac{2h}{32}\right)^{\frac{1}{2}}[/tex]
The expression can be simplified as:
[tex]t = 2\sqrt{\frac{h}{16}}[/tex]
For a player that has a hang time of 0.9s, the jump is found as follows:
[tex]0.9 = 2\sqrt{\frac{h}{16}}[/tex]
[tex]\sqrt{\frac{h}{16}} = \frac{0.9}{2}[/tex]
[tex](\sqrt{\frac{h}{16}})^2 = \left(\frac{0.9}{2}\right)^2[/tex]
[tex]h = 16\left(\frac{0.9}{2}\right)^2[/tex]
h = 3.24 feet.
For a player that has a hang time of 0.8s, the jump is found as follows:
[tex]0.8 = 2\sqrt{\frac{h}{16}}[/tex]
[tex]\sqrt{\frac{h}{16}} = \frac{0.8}{2}[/tex]
[tex](\sqrt{\frac{h}{16}})^2 = \left(\frac{0.8}{2}\right)^2[/tex]
[tex]h = 16\left(\frac{0.8}{2}\right)^2[/tex]
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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