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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(2h/32)^1/2, where t is the time in seconds, and h is the height of the jump, in feet.


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

Verified Answer :

Expression: t=2(2h/32)^1/2

Squaring on both sides, 
t² = 2(2h/32)
t² = 4h/32
t² = h/8
h = 8t²

Height of player 1 = 8(0.9)² = 6.48 Feet
Height of player 2 = 8(0.8)² = 5.12 Feet

Difference = 6.48 - 5.12= 1.36 Feet = 16.32 in

In short, Your Answer would be:16 Inches

Hope this helps!

The difference in jump of both players can be calculated by solving the given equation for \(h\).

The player-1 jump 8.00 inch higher than the player-2.

Given:

The hang time equation is \(t=2\left(\dfrac{2h}{32}\right)^\frac{1}{2}\)

Player 1 hang time is \(0.9\:\rm s\).

Player 2 hang time is \(0.8\: \rm s\).

Consider the given equation.

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

Now, solve for the \(h\).

\(t=2\left(\dfrac{2h}{32}\right)^\frac{1}{2}\\t=2\left(\dfrac{h}{16}\right)^\frac{1}{2}\\t=2\left(\dfrac{\sqrt{h} }{\sqrt{16}}\right)\\t=2\left(\dfrac{\sqrt{h} }{4}}\right)\\t=\left(\dfrac{\sqrt{h} }{2}}\right)\\h=2t^2\)

Calculate the height for Player 1.

\(h=2\times 0.9^2\\h=3.24 \:\rm feet\\h=38.88 \;\rm in\)

Calculate the height for Player 2.

\(h = 4(0.8)^2\\h= 2.56\:\rm feet\\h=30.72\:\ in\)

Calculate the difference in jump of both players,

\(= 38.88 - 30.72 = 8.16\)

Thus, the player-1 jump 8.00 inch higher than the player-2.

Learn more about height here:

https://brainly.com/question/12019427

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