IF a pro basketball player has a vertical hang-time of about 35 inches, what is his hang-time? Use hang-time function V=48T^2. This is how I have started this question
35=38T^With this my answer is 0.854, -0.854 Is this right?
35 inches is not the measurement of a "hang time". Please verify the wording of your question.
Is that the distance he is able to raise his center of mass?
What are the units of V in your equation?
I fear your teacher may be unqualified to teach the subject
check your typing, this looks weird
To find the hang-time of a pro basketball player given a vertical height of 35 inches, you need to rearrange the hang-time function V = 48T^2 and solve for T.
Starting with the equation 35 = 48T^2, divide both sides by 48 to isolate T^2:
35/48 = T^2
Now, take the square root of both sides to solve for T:
√(35/48) = √(T^2)
T ≈ ±0.71
However, it is important to note that time cannot be negative in this context. So the correct answer would be approximately T = 0.71 seconds.
Therefore, the hang-time of the basketball player is approximately 0.71 seconds.