Question 330514

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

His hang time is in how many seconds?
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V is given as 35 inches -- plug this into the equation and solve for t:
V=48t^2
35=48t^2
35/48 = t^2
sqrt(35/48) = t
0.85 secs = t