Question 249261
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)}}}
35={{{(48T^2)}}} [Substitute "35" for the V-value.  Solve for the T-value]
{{{(35/28)}}}={{{(48T^2/48)}}}
0.72916={{{(T^2)}}}
{{{(sqrt(0.72916))}}}={{{sqrt(T^2))}}}
0.8539 seconds=T