A basketball player has a vertical leap of about 36 in. If the equation d=48t^2 describes his hang time t as a function of his vertical leap d. what is his hang time? Show your work.

36=48t^2
t^2=3/4
t=(1/2)sqrt(3)seconds

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.