If a pro basketball player has a vertical leap of about 35 inches what is his hang-time? Use the hang-time function V=48T^2. Thank you I did mess up the typing last time didn't I?
If a pro basketball player has a vertical leap of about 35 inches, what is his hang-time? Using the hang-time function
V = 48T^2
I don't know how valid your equation is, but nevertheless, ...
35=48t^2
t^2 = .7291666..
t = ±.8539
but time is a positive number, so
his "hangtime" is 2(.854)units of time
or 1.7 units
(you don't say what t is)
No problem! To find the hang-time, we can rearrange the hang-time equation V = 48T^2 as T = sqrt(V/48).
Given that the basketball player has a vertical leap of 35 inches, we can substitute this value into the equation:
T = sqrt(35/48)
Calculating this, we find:
T ≈ sqrt(0.7292)
T ≈ 0.854 seconds
Therefore, the hang-time of the basketball player is approximately 0.854 seconds.
No problem! 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=48T^2, where V represents the vertical leap and T represents the hang-time.
We know that the vertical leap (V) is 35 inches. Let's substitute this value into the equation:
35 = 48T^2
To find the hang-time (T), we need to isolate T. So let's solve the equation step by step:
Divide both sides of the equation by 48:
35/48 = T^2
Now take the square root of both sides:
√(35/48) = T
Calculating this square root value, we get:
T ≈ 0.7604
Therefore, the hang-time of the pro basketball player with a vertical leap of about 35 inches is approximately 0.7604 seconds.