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

The amount of height that player 1 jumped higher than player 2 to the nearest inch is; 8 inches

We are given the hang time equation as;

t = 2(2h/32)^(½)

Where h is height of jump

Hang time of player 1 = 0.9 s

Hang time of player 2 = 0.8 s

Let's make h the subject of the hang time equation to get;

h = 2t²

For player 1;

h1 = 2(0.9)²

h1 = 3.24 ft

Converting to inches gives;

h1 = 38.88 inches

For player 2;

h2 = 2(0.8)²

h2 = 2.55 ft

Converting to inches gives

h2 = 30.72 inches

Difference in jump height = h1 - h2

Difference in jump height = 38.88 - 30.72

Difference in jump height ≈ 8 inches

Read more about hangtime equation at; https://brainly.com/question/3083779

Answer 2

Answer:

8 inches

Step-by-step explanation: