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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 play 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 high did player 1 jump than player 2

2 Answers

4 votes

Answer:

8 is the answer on edg

Explanation:

User JNYRanger
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6 votes

Answer:

The player-1 jump 0.68 feet higher than the player-2.

Explanation:

Given 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.

We have two players, player-1 and player-2. So the equations for each would be:- t₁=2√(2h₁/32) and t₂=2√(2h₂/32).

Then h₁ = 16(t₁/2)² and h₂ = 16(t₂/2)²

Given the hang time for each player, t₁ = 0.9 seconds and t₂ = 0.8 seconds.

h₁ = 16(0.9/2)² = 3.24 feet and h₂ = 16(0.8/2)² = 2.56 feet.

Finding the difference in jump of both players, h₁ - h₂ = 3.24 - 2.56 = 0.68 feet.

Hence, the player-1 jump 0.68 feet higher than the player-2.

User TheNavigat
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