A person standing close to the edge on top of a 72-foot building throws a ball vertically upward. The
quadratic function h = - 16ť+ 136t + 72 models the ball's height about the ground, h, in feet, t seconds after it was thrown.
a) What is the maximum height of the ball?
feet
b) How many seconds does it take until the ball hits the ground?
seconds

Answers

Answer 1

Answer:

I hope this helps.

Step-by-step explanation:

h(t) = -16t2 + 108t + 28

 

(a)

 

Maximum height occurs at the vertex of the height-vs.-time parabola, which is at

 

t = -108/[2(-16)] sec = ? sec

 

Evaluate h(t) at this value of t to get hmax.

 

(b)

 

Set h(t) = 0 and solve the quadratic equation for t.  You will get a positive and a negative solution.  Discard the negative solution, since time starts at t = 0.

Answer 2

The quadratic function, h = -16t + 136t + 72, gives us:

a) The maximum height of the ball is 360 feet.

b) The time taken by the ball to hit the ground is 9 seconds.

What is a quadratic function?

A quadratic function is a function over a quadratic expression, that is, over an expression having degree 2.

How to solve the question?

In the question, we are informed that a person standing close to the edge of a 72-foot building throws a ball vertically upward. The quadratic function h = -16t² + 136t + 72 models the ball's height above the ground, h in feet, t seconds after it was thrown.

We solve the following:-

a) What is the maximum height of the ball?

The maximum height of the ball can be obtained by differentiating the given quadratic function, h = -16t² + 136t + 72.

Differentiating both sides. we get:

[tex]\frac{\delta h}{\delta t} = -32t + 136[/tex] .

To get point of inflection, we equate this to 0, to get:

[tex]\frac{\delta h}{\delta t} = -32t + 136 = 0[/tex],

or, -32t + 136 = 0,

or, t = 136/32 = 4.5.

To check whether the value of h is maximum/minimum at t = 4.5, we differentiate [tex]\frac{\delta h}{\delta t} = -32t + 136[/tex] again, to get:

[tex]\frac{\delta^{2} h}{\delta t^{2} } = -32[/tex] , which is < 0, which that implies, h is maximum at t = 4.5.

Now, we calculate the maximum height, by putting in t = 4.5, in the equation, to get:

h = -16(4.5)² + 136(4.5) + 72,

or, h = -16(20.25) + 612 + 72,

or, h = -324 + 684 = 360.

Therefore, The maximum height of the ball is 360 feet.

b) How many seconds does it take until the ball hits the ground?

The time taken can be calculated by putting the value of h = 0, and solving the quadratic function h = -16t² + 136t + 72, as the height of the ball at the ground is 0.

Therefore, it can be written as:

0 = -16t² + 136t + 72.

or, 16t² - 136t - 72 = 0.

Solving this using the quadratic equation, we get:

[tex]t = \frac{-(-136)\pm \sqrt{(-136)^{2} - 4(16)(-72)}}{2(16)}[/tex] ,

or, [tex]t = \frac{136\pm \sqrt{18496 + 4608}}{32}[/tex] ,

or, [tex]t = \frac{136\pm \sqrt{23104}}{32}[/tex] ,

or, [tex]t = \frac{136\pm 52}{32}[/tex]

Therefore, either t = (136+152)/32 = 9,

or, t  (136 - 152)/32 = -0.5.

Since t represents time, we won't take the negative value, and hence t = 9.

Therefore, The time taken by the ball to hit the ground is 9 seconds.

Learn more about quadratic equations at

https://brainly.com/question/15340389

#SPJ2


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Answers

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Step-by-step explanation:

I could be wrong

The usual speed of Eric's mother is 4 miles per hour .

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