Elaine shoots an arrow upward at a speed of 32 feet per second from a bridge that is 28 feet high. The height of the arrow is given by the function h(t) = -16t2+32t + 28, where t is the time in seconds.
a. What is the maximum height that the arrow reaches?
b. How long does it take the arrow to reach its maximum height?
c. How long would it take before the arrow reached the ground? Round your answer to the hundredths place.
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a. To find the maximum height that the arrow reaches, we need to find the vertex of the parabolic function h(t) = -16t^2 + 32t + 28. The vertex of a parabola can be found using the formula t = -b / (2a), where a, b, and c are the coefficients of the quadratic function. In this case, a = -16 and b = 32, so the formula becomes t = -32 / (2*-16). Simplifying it further, t = -32 / -32 = 1.
To find the maximum height, we substitute t = 1 into the function h(t): h(1) = -16(1)^2 + 32(1) + 28 = -16 + 32 + 28 = 44 feet.
Therefore, the maximum height that the arrow reaches is 44 feet.
b. The time taken for the arrow to reach its maximum height is equal to the x-coordinate of the vertex, which we found to be t = 1 second.
Therefore, it takes 1 second for the arrow to reach its maximum height.
c. To find the time it takes for the arrow to reach the ground, we need to find the point at which its height is 0. We can solve this by setting the function h(t) = 0 and solving for t.
0 = -16t^2 + 32t + 28
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not convenient, so we'll use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a).
Using the values from our equation, a = -16, b = 32, and c = 28, we can substitute them into the formula: t = (-32 ± √(32^2 - 4(-16)(28))) / (2(-16)).
Simplifying further, t = (-32 ± √(1024 + 1792)) / (-32).
t = (-32 ± √(2816)) / (-32).
Now we calculate the square root of 2816, which is approximately 53.04.
t = (-32 ± 53.04) / (-32).
Now we calculate the two possible values of t:
t1 = (-32 + 53.04) / -32 ≈ -21.04 / -32 ≈ 0.6575 seconds.
t2 = (-32 - 53.04) / -32 ≈ -85.04 / -32 ≈ 2.658 seconds.
Since time cannot be negative, we discard t1 ≈ 0.6575. Therefore, the arrow takes approximately 2.66 seconds to reach the ground.
So, it would take approximately 2.66 seconds before the arrow reached the ground.