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

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

Step-by-step 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.

8 is the answer on edg