The height hh (in feet) of a volleyball tt seconds after it is hit can be modeled by h = -16t^2+48t+4h=16t2+48t+4. How long is the volleyball in the air?

1 Answer
Nov 15, 2015

t = 3.08st=3.08s

Explanation:

We have to substitute hh by 00, as that is the height both when projected as when it hits the ground. Then we solve for tt:
-16t^2 + 48t + 4 = 016t2+48t+4=0,
16t^2 - 48t - 4 = 016t248t4=0,
4(4t^2 - 12t - 1) = 04(4t212t1)=0,
4t^2 - 12t - 1 = 04t212t1=0. Now we use t = (-b +- sqrt((b^2 - 4ac)))/(2a)t=b±(b24ac)2a:
t = (- (-12) +- sqrt((-12)^2 - 4(4)(-1)))/(2(4))t=(12)±(12)24(4)(1)2(4),
t = (12 +- sqrt(144 + 16))/(8)t=12±144+168,
t = (12 +- sqrt(160))/(8)t=12±1608,
t = 3.08st=3.08s.

Hope it Helps! :D .