SOLUTION: 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.
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. Found 3 solutions by ewatrrr, stanbon, CharlesG2:Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Hi
-16t^2 +560t = 0
-16t(t - 35)= 0
t = 0 or t = 35sec
35sec is when the object hits the ground
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website! h(t) = -16t^2 +560t
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h(t) is the height of the object after t seconds.
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When the object is on the ground the height is zero.
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Solve:
-16t^2+560t = 0
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-16t(t-35) = 0
t = 0 or t = 35
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That means the object is on the ground at the beginning
and it is back on the ground after 35 seconds.
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Cheers,
Stan H. Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website! 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