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)=−16t^2+8t+24
Include all necessary work.
a) Find the maximum height that the acorn reaches. After how many seconds does the acorn reach this maximum height?
b) How many seconds would it take the acorn to reach the ground?
a) To find the maximum height that the acorn reaches, we need to find the vertex of the parabola described by the equation h(t) = -16t^2 + 8t + 24.
The x-coordinate of the vertex can be found using the formula x = -b/2a, where a = -16 and b = 8.
x = -8/(2(-16)) = -8/-32 = 1/4
To find the y-coordinate of the vertex, substitute this value of t into the equation:
h(1/4) = -16(1/4)^2 + 8(1/4) + 24
h(1/4) = -4 + 2 + 24
h(1/4) = 22
Therefore, the maximum height that the acorn reaches is 22 feet.
To find the time it takes for the acorn to reach this maximum height, we take the x-coordinate of the vertex, which is 1/4 seconds.
b) To find the time 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
Dividing the equation by -8, we get:
0 = 2t^2 - t - 3
This equation can be factored as:
0 = (2t + 3)(t - 1)
Setting each factor equal to zero, we find:
2t + 3 = 0 or t - 1 = 0
Solving these equations, we get:
t = -3/2 or t = 1
Since time cannot be negative, the acorn takes 1 second to reach the ground.