A player bumps a volleyball with an initial vertical velocity of 20 ft/s. The height of the ball can be model by the function h(t) = -16t2 + 20t + 4, where h is the height in feet and t is the time in seconds.

a. What is the maximum height of the ball? ( use your graphing calculator or find the vertex)

d. Suppose the volleyball were hit under the same conditions, but with an initial velocity of 32 ft/s. How much higher would the ball go?

What I’m the world. help with the question.

your equation should have been

h(t) = -16t^2 + 20t + 4

a) It is your graphing calculator, I don't know what you have nor do I know how to use yours.
Follow the steps outlined in the manual to answer a)

OR

for f(x) = ax^2 + bx + c
the x of the vertex is -b/(2a)
so for yours, the t of the vertex is -20/(-32) = .625
h(.625) = -16(.625^2) + 20(.625) + 4 = 10.25
so the vertex is (.625, 10.25)
then the maximum height is 10.25 ft.

OR

h '(t) = -32t + 20 = 0 for a max/min of h(t)
32t = 20
t = .625 as above, find h(.625)

OR
complete the square, a method you might have learned in your course.

b) change 20t to 32t in your equation and repeat a)

To find the maximum height of the ball, we can find the vertex of the given quadratic function. The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the formula x = -b/2a.

a. In this case, the function is h(t) = -16t^2 + 20t + 4. Comparing this with the general form, we can see that a = -16 and b = 20.

Using the formula x = -b/2a, we can find the time at which the ball reaches its maximum height:

t = -20 / (2 * (-16))
t = -20 / (-32)
t = 0.625 seconds

Now, substitute this value of t back into the equation to find the maximum height:

h(0.625) = -16(0.625)^2 + 20(0.625) + 4
h(0.625) = -6.25 + 12.5 + 4
h(0.625) = 10.25 feet

So, the maximum height of the ball is 10.25 feet.

d. Now, let's consider the situation where the initial velocity is 32 ft/s. In this case, the function is still h(t) = -16t^2 + 20t + 4, but the initial velocity term changes to 32t.

We can follow the same steps as before to find the time at which the ball reaches its maximum height:

t = -20 / (2 * (-16))
t = -20 / (-32)
t = 0.625 seconds

The difference now is that the initial velocity term is 32t, so the new equation for the height is:

h(t) = -16t^2 + 32t + 4

Substituting the value of t back into the equation, we can find the new maximum height:

h(0.625) = -16(0.625)^2 + 32(0.625) + 4
h(0.625) = -6.25 + 20 + 4
h(0.625) = 17.75 feet

Therefore, if the initial velocity is 32 ft/s instead of 20 ft/s, the ball would go approximately 7.5 feet higher.

To find the maximum height of the ball, we need to find the vertex of the function h(t) = -16t^2 + 20t + 4.

a. The vertex of a quadratic function can be found using the formula t = -b/2a, where a, b, and c are the coefficients of the quadratic function in the form ax^2 + bx + c.

In this case, a = -16 and b = 20. Plugging these values into the formula, we have:

t = -(20) / (2*(-16))
t = -20 / (-32)
t = 5/8

To find the maximum height, substitute this value back into the function h(t):

h(5/8) = -16(5/8)^2 + 20(5/8) + 4
h(5/8) = -16(25/64) + 100/8 + 4
h(5/8) = -400/64 + 800/64 + 256/64
h(5/8) = 656/64

Simplifying, we get:

h(5/8) = 41/4
h(5/8) = 10.25 ft

So, the maximum height of the ball is 10.25 ft.

d. Now let's find the maximum height when the initial velocity is 32 ft/s. We can use the same process of finding the vertex as before.

In this case, a = -16 and b = 32. Using the formula t = -b/2a:

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

Substituting t = 1 back into the function h(t):

h(1) = -16(1)^2 + 32(1) + 4
h(1) = -16 + 32 + 4
h(1) = 20

So, when the initial velocity is 32 ft/s, the ball reaches a maximum height of 20 ft.

To find how much higher the ball goes, we can subtract the maximum height in the first case (10.25 ft) from the maximum height in the second case (20 ft):

20 ft - 10.25 ft = 9.75 ft

Therefore, the ball goes 9.75 feet higher when the initial velocity is increased from 20 ft/s to 32 ft/s.