SOLUTION: if an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t seconds is given by the equation h = -16t^2+vt+s, wh

Question 811813: if an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t seconds is given by the equation h = -16t^2+vt+s, where h is in feet. If the object is propelled from a height of 12 feet with an initial velocity of 96 feet per second, its height h is given by the equation h = 16t^2+64t+8. After how many seconds is the height is 120 feet?
Found 2 solutions by josmiceli, Alan3354:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Since this is a parabola, there will likely
be 2 solutions. One solution is the time
when the object is on it's way up and reaches
120 ft. The other solution is the time when
the object is on it's way down and reaches
120 ft again.
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( actually, it looks like the 8 should be a 12,
which is the initial height, but I'll leave it 8 )

I'll complete the square

Both sides need to be a perfect square for this
to work, and is not a square of anything.
I don't think the object ever reaches 120 ft
Here's a plot:

It looks like it's maximum height is 72 ft at 2 sec
Is something copied wrong?
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Note: it's = it is

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
if an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t seconds is given by the equation h = -16t^2+vt+s, where h is in feet. If the object is propelled from a height of 12 feet with an initial velocity of 96 feet per second, its height h is given by the equation h = 16t^2+64t+8. After how many seconds is the height is 120 feet?
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The text says 96 ft/sec but the equation says 64 ft/sec
And the launch heights are different, 12 and 8 feet.
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