1. First, allow me to verity your equation, using physics.
The actual hang time function is, if V is the vertical distance in feet,
V = (1/2)g(t/2)^2
= (1/8)*32.2*t^2 = 4.025 t^2
or 48.3 t^2, if V is in inches.
So, your equation is OK
For your question, use T = sqrt(V/48)
= sqrt(20/48) = 0.645 seconds
2. Let y = x^2, and solve
10y^2 -11y + 3 = 0
(5y - 3)(2y -1) = 0
y = 3/5 or 1/2
x = +/-sqrt(3/5) or +/-sqrt(1/2)
or, in decimal form,
0.7746..
-0.7746..
0.7071..
-0.7071
1. If a pro basketball player has a vertical leap of about 20 inches, what is his hang time? Use the hang-time function V=48 T^2
2.10x^4-11x^2+3=0
1 answer
1 answer