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.
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.
Your original answer of t=5.679 or 1.321 seconds is incorrect. The correct times are t=2.5 seconds on the way up and t=3.5 seconds on the way down.
To find the time it takes for the arrow to reach a height of 180 ft, you can set up the equation:
180 = -16t^2 + 112t
Rearranging the equation, we get:
16t^2 - 112t + 180 = 0
This is a quadratic equation, which can be solved using the quadratic formula:
t = [-b ± √(b^2 - 4ac)] / 2a
Plugging in the values from our equation, we get:
t = [112 ± √((-112)^2 - 4(16)(180))] / (2 * 16)
Simplifying further:
t = [112 ± √(12544 - 11520)] / 32
t = [112 ± √(1024)] / 32
Since we're looking for the times, we take both the positive and negative solutions:
t1 = (112 + √1024) / 32
t2 = (112 - √1024) / 32
Evaluating these expressions, we get:
t1 = (112 + 32) / 32 = 144 / 32 = 4.5 seconds
t2 = (112 - 32) / 32 = 80 / 32 = 2.5 seconds
Therefore, the arrow takes 2.5 seconds to reach a height of 180 ft. on the way up, and 4.5 seconds to reach the same height on the way down.
Your original answer of t = 5.679 or 1.321 s is incorrect. Here is the correct solution:
To find the time it takes for the arrow to reach a height of 180 ft, we can set up the equation:
180 = -16t^2 + 112t
Rearranging the equation, we get:
16t^2 - 112t + 180 = 0
This is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = [-b +/- sqrt(b^2 - 4ac)] / 2a
In this case, a = 16, b = -112, and c = 180:
t = [-(-112) +/- sqrt((-112)^2 - 4(16)(180))] / (2*16)
= [112 +/- sqrt(12544 - 11520)] / 32
= [112 +/- sqrt(1024)] / 32
= [112 +/- 32] / 32
So we have two possible solutions:
t1 = (112 + 32) / 32
= 144 / 32
= 4.5 s
t2 = (112 - 32) / 32
= 80 / 32
= 2.5 s
Therefore, the arrow takes 4.5 seconds to reach a height of 180 ft when it is on its way down, and it takes 2.5 seconds to reach the same height on its way up.