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) = −16t2 + 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

the max height is achieved at the vertex of the parabola, when t = 1

for (c), just solve

16t^2 - 32t - 240 = 0
16(t+3)(t-5) = 0
...

To find the maximum height of the projectile, we need to determine the vertex of the function h(t) = -16t^2 + 32t + 240. The vertex of a quadratic function, in the form of h(t) = at^2 + bt + c, can be found using the formula t = -b / (2a).

In our case, a = -16 and b = 32. Substituting these values into the formula, we get:

t = -b / (2a)
t = -32 / (2 * -16)
t = -32 / -32
t = 1

The time t when the projectile achieves its maximum height is 1 second.

To find the maximum height, substitute the value of t = 1 into the function h(t):

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

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

To find the time t when the projectile has a height of 0 feet, we need to solve the equation h(t) = 0. Using the quadratic formula, we can find the values of t:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation h(t) = -16t^2 + 32t + 240, a = -16, b = 32, and c = 240. Substituting these values into the quadratic formula, we get:

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

Given that we have a plus-minus sign, we have two possible solutions:

When t = (-32 + 128) / -32 = 96 / -32 = -3
When t = (-32 - 128) / -32 = -160 / -32 = 5

However, since time cannot be negative in this context, the only valid solution is t = 5 seconds when the projectile reaches a height of 0 feet.