Use the following information to answer each part.
An acorn is tossed upwards with an initial velocity of 8 ft/sec from the top of a 24-foot building. The model given describes the height, h, of the acorn as a function of time, t, in seconds.
h(t)=−16t2+8t+24
Make sure to include all necessary work in order to receive full credit.
a) Find the maximum height that the acorn reaches. After how many second does the acorn reach this maximum height?
b) How many seconds would it take the acorn to reach the ground?
(5 points)
a) To find the maximum height that the acorn reaches, we need to find the vertex of the parabolic function h(t) = -16t^2 + 8t + 24. The vertex of a parabola is given by the formula (t, h), where t is the x-coordinate and h is the y-coordinate.
The x-coordinate of the vertex is given by the formula t = -b/2a, where a, b, and c are the coefficients of the function. In this case, a = -16 and b = 8, so t = -8/(2*(-16)) = -8/(-32) = 1/4.
Substituting t = 1/4 into the function, we can find the y-coordinate of the vertex:
h(t) = -16(1/4)^2 + 8(1/4) + 24
= -16(1/16) + 2 + 24
= -1 + 2 + 24
= 25
Therefore, the maximum height that the acorn reaches is 25 feet. It reaches this height after 1/4 seconds.
b) To find how many seconds it takes for the acorn to reach the ground, we need to find the value of t when h(t) = 0.
0 = -16t^2 + 8t + 24
We can solve this equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from the equation: a = -16, b = 8, and c = 24, we have:
t = (-8 ± √(8^2 - 4(-16)(24))) / (2(-16))
= (-8 ± √(64 + 1536)) / (-32)
= (-8 ± √1600) / (-32)
= (-8 ± 40) / (-32)
Simplifying, we have two possible solutions:
t1 = (-8 + 40) / (-32) = 32 / (-32) = -1
t2 = (-8 - 40) / (-32) = -48 / (-32) = 3/2
Since time cannot be negative in this context, the acorn reaches the ground after 3/2 seconds, or 1.5 seconds.