Use the function V(T) = 18T2. A basketball player has a vertical leap of 38.5 inches.
What is their hang time to the nearest hundredth of a second?
Hint: T is your time in seconds.
If V = 18 T^2 ,
with V (the vertical height) in inches and T in seconds, then
T = sqrt(V/18)
If V = 38.5 inches, then
T = sqrt(38.5/18) = 1.463 seconds
"sqrt" means "square root of"
Thank you so much for helping me.I appreciate it
I must admit that I am caught totally unaware of the recent change in the law of gravity.
I think drwls did not see the inches unit.
I thought it was
height = 16t^2 , where height is in feet
so 38.5 = 38.5/12 ft = 3.208 ft
3.208 = 16t^2
t = .2005
t = √.2005 = .448 seconds to go up
so hang time = 2(.448) or appr 0.9 seconds
(Michael Jordan was timed at a 1 second hangtime, I looked it up)
To find the hang time of the basketball player, we need to solve the equation V(T) = 18T^2 for T.
Here, V(T) represents the vertical velocity at time T, and 18T^2 represents the given function.
Given that the basketball player has a vertical leap of 38.5 inches, we can rewrite the equation as:
38.5 = 18T^2
To solve for T, we need to isolate T^2. Divide both sides of the equation by 18:
38.5/18 = T^2
Simplifying the equation:
2.14 ≈ T^2
To find T, we need to take the square root of both sides of the equation. However, we must be careful here because the equation has two possible answers: the positive square root and the negative square root.
Taking the square root of both sides, we have:
T = ±√2.14
Since time cannot be negative in this context, we can discard the negative square root. Therefore:
T ≈ √2.14
Using a calculator, we can find the approximate value of √2.14:
T ≈ 1.46
Therefore, the hang time, rounded to the nearest hundredth of a second, is approximately 1.46 seconds.