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.