If a pro basketball player has a vertical leap of about 35 inches, what is his hang time? Use the hang-time function V = 48T²
T = V/48 = 35 / 48 = 0.73S
T^2 = 35/48 = 0.729,
T = 0.854s.
To find the hang time, we need to use the hang-time function V = 48T², where V represents the vertical leap and T represents the hang time.
Given that the vertical leap of the basketball player is 35 inches, we can substitute V as 35 into the equation:
35 = 48T²
To solve for T, we need to isolate T on one side of the equation.
Divide both sides of the equation by 48 to get:
35/48 = T²
T² ≈ 0.729
To find T, we take the square root of both sides of the equation:
√(T²) ≈ √0.729
T ≈ √0.729
T ≈ 0.854
Therefore, the hang time of the basketball player is approximately 0.854 seconds.