The height in feet of a projectile with an initial velocity of 32 feet per second and an initial height of 240 feet is a function of time in seconds given by

h(t) = −16t^2 + 32t + 240.

(a) Find the maximum height of the projectile. ______ ft

(b) Find the time t when the projectile achieves its maximum height.
t = ______ sec

(c) Find the time t when the projectile has a height of 0 feet.
t = ______ sec

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To find the maximum height of the projectile, we need to determine the vertex of the quadratic function h(t) = -16t^2 + 32t + 240.

(a) The maximum height of the projectile corresponds to the y-coordinate of the vertex. The x-coordinate of the vertex represents the time at which the projectile reaches its maximum height.

To find the x-coordinate of the vertex, we use the formula x = -b / (2a), where a = -16 and b = 32 in our case.

x = -32 / (2 * -16)
x = -32 / -32
x = 1

So, the time when the projectile achieves its maximum height is t = 1 sec.

Now, we substitute t = 1 sec into the function h(t) to find the maximum height:

h(t) = -16(1)^2 + 32(1) + 240
h(t) = -16 + 32 + 240
h(t) = 256

Therefore, the maximum height of the projectile is 256 ft.

(b) We have already determined that the time t when the projectile achieves its maximum height is 1 sec.

(c) To find the time t when the projectile has a height of 0 ft, we set h(t) = 0 and solve for t.

0 = -16t^2 + 32t + 240

To solve this quadratic equation, we can factorize it or use the quadratic formula. In this case, let's use the quadratic formula:

t = [-b ± sqrt(b^2 - 4ac)] / (2a)

Here, a = -16, b = 32, and c = 240.

t = [-32 ± sqrt((32)^2 - 4(-16)(240))] / (2 * -16)
t = [-32 ± sqrt(1024 + 15360)] / (-32)
t = [-32 ± sqrt(16384)] / (-32)
t = [-32 ± 128] / (-32)

Using both the positive and negative square root solutions, we get two possible times:

t = (-32 + 128) / (-32)
t = 96 / (-32)
t = -3

t = (-32 - 128) / (-32)
t = -160 / (-32)
t = 5

Therefore, the time when the projectile has a height of 0 ft is t = -3 sec and t = 5 sec.