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To find the hang time of the basketball player, we'll start by rearranging the given equation:
d = 48t^2
Given that the vertical leap of the player is 36 inches, we can substitute d with 36:
36 = 48t^2
Now, let's solve for t by isolating the variable term (t^2):
36/48 = t^2
3/4 = t^2
Next, we need to get rid of the square on t. To do this, we can take the square root of both sides of the equation:
√(3/4) = √(t^2)
Simplifying the square root, we have:
√3/√4 = t
√3/2 = t
So, the hang time of the basketball player is √3/2 seconds or (1/2)√3 seconds.