Question

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Answer 1

The maximum height of the rocket is 43.89 feet

How to write the function

The general function is given as:

h(t) = -16t^2 + vt + h

The initial velocity is

v = 53

So, we have:

h(t) = -16t^2 + 53t + h

The initial height is

h = 0

So, we have:

h(t) = -16t^2 + 53t

Hence, the function of the height is h(t) = -16t^2 + 53t

The maximum height of the rocket

In (a), we have:

h(t) = -16t^2 + 53t

Differentiate the function

h'(t) = -32t + 53

Set to 0

-32t + 53 = 0

This gives

-32t = -53

Divide by -32

t = 1.65625

Substitute t = 1.65625 in h(t) = -16t^2 + 53t

h(1.65625) = -16 * 1.65625^2 + 53 * 1.65625

Evaluate

h(1.65625) = 43.890625

Approximate

h(1.65625) = 43.89

Hence, the maximum height of the rocket is 43.89 feet

Time to hit the ground

In (a), we have

h(t) = -16t^2 + 53t

Set to 0

-16t^2 + 53t = 0

Divide through by -t

16t - 53 = 0

Add 53 to both sides

16t = 53

Divide by 16

t = 3.3125

Hence, the time to hit the ground is 3.3125 seconds

The graph of the function h(t)

See attachment for the graph of the function h(t)

Read more about height functions at:

brainly.com/question/12446886

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