Question 396663
How do I find when an object hits the ground using the equation: -16t^2 +560t.  I know how to find the height after however many seconds, but cannot figure out how to find this.


h = -16t^2 + 560t
when hits the ground: set h to 0
0 = -16t^2 + 560t
0 = t^2 - 35t (35 * 16 = 560)
(-35/2)^2 = t^2 - 35t + (-35/2)^2 (completed the square)
(-35/2)^2 = (t - 35/2)^2 (by FOIL, First Outer Inner Last, (-35/2)t * 2 = -35t are 2 middle terms added)
-35/2 = t - 35/2 or -35/2 = -t + 35/2
0 = t ........... or -70/2 = -t
...................... 70/2 = 35 = t
check with t = 35:
h = -16(35)^2 + 560 * 35
h = -16 * 1225 + 19600
h = -19600 + 19600
h = 0, yes
answer: in 35 seconds