Your original answers were close but slightly off. The correct answer should be:
The arrow reaches a height of 180 ft at two different times: t = 2.5 seconds on its way up and t = 3.5 seconds 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
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?
Your two answers are those two times
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.
1 answer