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.
step 1: this is a parabola, find the two zeroes.
0=-16t^2+32t+28
0=-(8t^2-16t-14)
=-(4t^2-8t-7)
going to the quadratic formula
t=(-b+-sqrt(b^2-4ac))/2a
a=4; b=-8, c=-7
t=(8+-sqrt(64+4*56)/8
t= 1+-16.1/8= 3 or t=-1
so where is the maximum height? Halfway between..t=(-1+3)/2 or at 1 second
max height=-16t2+32t + 28 at t=1 solve it
when will it hit the ground? t=3 as above
To find the maximum height reached by the arrow, we need to determine the vertex of the quadratic function. The vertex of a quadratic function in the form ax^2 + bx + c can be found using the formula x = -b/2a.
In this case, the function is h(t) = -16t^2 + 32t + 28. Comparing this to the general form, we have a = -16 and b = 32.
Using the formula x = -b/2a, we can calculate the time it takes for the arrow to reach its maximum height:
t = -32 / (2 * -16)
t = -32 / -32
t = 1
Therefore, it takes 1 second for the arrow to reach its maximum height.
To find the maximum height, we substitute the value of t into the function h(t):
h(1) = -16(1)^2 + 32(1) + 28
h(1) = -16 + 32 + 28
h(1) = 44
The maximum height reached by the arrow is 44 feet.
To determine how long it takes for the arrow to reach the ground, we need to find the time when the height is equal to zero. We can solve this by setting the function h(t) equal to zero and solving for t:
-16t^2 + 32t + 28 = 0
To solve this equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a). In this case, a = -16, b = 32, and c = 28.
t = (-32 ± √(32^2 - 4(-16)(28))) / (2(-16))
t = (-32 ± √(1024 + 1792)) / (-32)
t = (-32 ± √2816) / (-32)
Since we are only interested in the positive value for t, we take the positive square root:
t = (-32 + √2816) / (-32)
Using a calculator, we find that √2816 ≈ 53.09:
t = (-32 + 53.09) / (-32)
t ≈ 1.66
Rounding to the hundredths place, it would take approximately 1.66 seconds for the arrow to reach the ground.
To find the maximum height that the arrow reaches, we need to determine the vertex of the parabolic equation.
a. The height of the arrow is given by the function h(t) = -16t^2 + 32t + 28, where t is the time in seconds. The vertex form of a quadratic equation is given by h(t) = a(t - h)^2 + k, where (h, k) represents the vertex of the parabola.
Comparing this form with the given equation, we have:
a = -16
h = -b/2a = -32 / (2 * -16) = 1
k = h(1) = -16(1)^2 + 32(1) + 28 = -16 + 32 + 28 = 44
Therefore, the maximum height that the arrow reaches is 44 feet.
b. To find the time it takes for the arrow to reach its maximum height, we need to determine the value of t when h(t) is maximum. Since the vertex is the maximum point of the parabola, we can say that t = h.
Thus, the arrow reaches its maximum height after 1 second.
c. To find when the arrow reaches the ground, we need to determine the time it takes for h(t) to be zero (ground level). So, we need to solve the quadratic equation h(t) = 0.
Since h(t) = -16t^2 + 32t + 28 = 0, we can solve this equation by factoring (if possible) or by using the quadratic formula.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a)
a = -16, b = 32, c = 28
t = ( - 32 ± √(32^2 - 4 * -16 * 28)) / (2 * -16)
t = ( - 32 ± √(1024 + 1792)) / -32
t = ( - 32 ± √2816) / -32
Using a calculator, we can find the exact values of the square root. The result is approximately t = 1.61, -0.11.
Since the time cannot be negative in this context, we only consider the positive value, t ≈ 1.61.
Therefore, it would take approximately 1.61 seconds before the arrow reaches the ground, rounding to the hundredths place.
(b) as always, max height (the vertex) is at t = -b/2a = 1
(a) figure h(1)
(c) solve -16t^2+32t+28 = 0