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
Distance, V, in inches travelled from rest under gravity
= (1/2)gt²
= (1/2)32'*12"/' t²
Since the hang time T includes upward and downward movements, T=2t, so equation above becomes
192(T/2)² = V
hence
V=48T²
Substitute V=35" to get
35=48T²
Solve to get
T=sqrt(35/48)=0.854 sec.
To find the hang time of a pro basketball player with a vertical leap of about 35 inches, we can use the hang-time function V = 48T², where V represents the vertical leap in inches and T represents the hang time in seconds.
1. Start by converting the vertical leap from inches to feet. Since there are 12 inches in a foot, divide 35 inches by 12:
35 inches / 12 = 2.92 feet
2. Plug the value of the vertical leap (V) into the hang-time function:
2.92 = 48T²
3. Divide both sides of the equation by 48 to isolate T²:
2.92 / 48 = T²
0.06083 = T²
4. Take the square root of both sides to solve for T:
√(0.06083) = T
T ≈ 0.247 seconds
Therefore, the hang time of the basketball player with a vertical leap of about 35 inches is approximately 0.247 seconds.