# math,correction

Can someone correct my answers thanks.

Problem #1
The equation h= -16t^2+112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a height of 180 ft.

My answer: t = 5.679 or 1.321 s

Yes, and you ought to know what the arrow is doing at each time. How can it be at that height twice?

I have no idea....that is what i end up with. what should i do?

jasmine, wouldn't the arrow reach the stated height on its way up and then again on its way down?

Can someone correct my answers thanks.

Problem #1
The equation h= -16t^2+112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a height of 180 ft.

My answer: t = 5.679 or 1.321 s

The time it takes to reach its maximum height derives from

Vf = Vo - gt
0 = 112 - 32t
t = 3.5 sec.

The maximum height reached is
h = Vot - 16t
h = 112(3.5) - 16(3.5)^2
h = 392 - 192 = 196 ft.

180 = -16t^2 + 112t
4t^2 -28t + 45 = 0
t = [28+/-sqrt(28^2 - 4(4)45)]/8
t = [28+/-8]/8
t = 2.5 sec. on the way up and 3.5 sec. onthe way down.

Alternatively:
On the way down it falls 16ft. to reach 180 ft. again.

Therefore, 16 = (0)t + 16t^2 or
t = 1 sec. + 2.5 = 3.5 sec. on the way down.

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