If a pro basketball player has a vertical leap of about 25 inches, what is his hang time? Use the hang-time function V=48T^
I am having major problems here!!! I need help.
V = 48T^2,
25 = 48T^2,
T^2 = 25/48 = 0.52s.
T = 0.72s.
To calculate the hang time of a pro basketball player with a vertical leap of 25 inches, you need to use the hang-time function V=48T^2.
The hang time equation tells us that the vertical leap (V) is equal to 48 times the hang time (T) squared. In this case, the vertical leap is 25 inches.
To find the hang time, we need to rearrange the equation to solve for T. Here's how you can do it:
1. Start with the equation V=48T^2.
2. Divide both sides of the equation by 48: V/48 = T^2.
3. Take the square root of both sides of the equation: sqrt(V/48) = T.
Now, let's substitute the value of the vertical leap (V = 25 inches) into the equation to find the hang time:
T = sqrt(25/48)
Using a calculator, you can find the square root and calculate the hang time:
T ≈ 0.382 seconds
Therefore, the hang time of the pro basketball player with a vertical leap of 25 inches is approximately 0.382 seconds.