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? in.

Answer 1

Player 1 jumps 0.68in higher than Player 2

Given the equation for hang time expressed as:

• $t=2(2h/32)^{1/2}$

Square both sides of the equation to have:

$t^2 = 4(2h/32)\\t^2 = 8h/32\\t^2 = 0.25h\\h = \frac{t^2}{0.25}$

If the time of jump for player 1 is 0.9s, the height covered will be "

h = 0.9²/0.25

h = 0.81/0.25

h = 3.24 in

If the time of jump for player 2 is 0.8s, the height covereed will be "

h = 0.8²/0.25

h = 0.64/0.25

h = 2.56 in

Difference in height = 3.24in - 2.56in

Difference in height = 0.68in

Hence Player 1 jumps 0.68in higher than Player 2

Leearn more on subject of formula here: https://brainly.com/question/657646