If a pro basketball player has a vertical leap of about 20in., what is his hang time?
Use the hang time function v=49t^2
V = 49t^2,
t^2 = V/49,
t^2 = 20 / 49 = 0.408s.
To find the hang time of a basketball player, we can use the given hang time function v = 49t^2 where v represents the vertical leap and t represents the hang time.
Given that the vertical leap of the basketball player is about 20 inches, we can substitute this value into the equation and solve for t.
20 = 49t^2
To isolate t, we first divide both sides of the equation by 49:
20/49 = t^2
Now, to solve for t, we need to take the square root of both sides of the equation:
sqrt(20/49) = t
This simplifies to:
t ≈ 0.428 seconds (rounded to three decimal places)
Therefore, the hang time of the basketball player is approximately 0.428 seconds.