Question 226395
The problem gives you:
v=48t^2
where
v is "vertical leap" (in inches)
t is time (in seconds)
.
Plug in the given "vertical leap" of 20 inches and solve for t:
v=48t^2
20=48t^2
20/48 = t^2
{{{sqrt(20/48) = t}}}
0.645 seconds = t (hang time)