Asked by deja
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.
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First simplify the formula:
t = 2√(2h/32) = 2√(h/16) = √h / 2, so √h = 2t and h = 4t^2.
Compute heights:
- Player 1: h1 = 4(0.9)^2 = 4(0.81) = 3.24 ft
- Player 2: h2 = 4(0.8)^2 = 4(0.64) = 2.56 ft
Difference: 3.24 − 2.56 = 0.68 ft = 0.68·12 = 8.16 in ≈ 8 in.
Answer: 8 in.
t = 2√(2h/32) = 2√(h/16) = √h / 2, so √h = 2t and h = 4t^2.
Compute heights:
- Player 1: h1 = 4(0.9)^2 = 4(0.81) = 3.24 ft
- Player 2: h2 = 4(0.8)^2 = 4(0.64) = 2.56 ft
Difference: 3.24 − 2.56 = 0.68 ft = 0.68·12 = 8.16 in ≈ 8 in.
Answer: 8 in.
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