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

1 Answer
Hugo C.
Nov 15, 2015

#t = 3.08s#

Explanation:

We have to substitute #h# by #0#, as that is the height both when projected as when it hits the ground. Then we solve for #t#:
#-16t^2 + 48t + 4 = 0#,
#16t^2 - 48t - 4 = 0#,
#4(4t^2 - 12t - 1) = 0#,
#4t^2 - 12t - 1 = 0#. Now we use #t = (-b +- sqrt((b^2 - 4ac)))/(2a)#:
#t = (- (-12) +- sqrt((-12)^2 - 4(4)(-1)))/(2(4))#,
#t = (12 +- sqrt(144 + 16))/(8)#,
#t = (12 +- sqrt(160))/(8)#,
#t = 3.08s#.

Hope it Helps! :D .

Related questions
Impact of this question
5923 views around the world
You can reuse this answer
Creative Commons License