Question

1. \\(4x^2 - 12x + 7 = 0\\)\

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problems 8 - 10: use the quadratic formula to answer each question. show work.\
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1. the height \\(h(t)\\) in feet of an object \\(t\\) seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation \\(h(t) = -16t^2 + 60t\\). at what times will the object be at a height of 56 feet?\

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1. the formula \\(d = 0.05s^2 + 1.1s\\) estimates the minimum stopping distance \\(d\\) in feet for a car traveling \\(s\\) miles per hour. if a car stops in 200 feet, what is the fastest it could have been traveling when the driver applied the brakes? round your answer to the nearest tenth, and label.\

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1. in 1990, american randy barnes set the world record for the shot put. his throw can be described by the equation \\(y = -16x^2 + 368x\\), where \\(x\\) is the number of horizontal meters the shot put has gone and \\(y\\) is the number of meters that the shot put is above the ground. use the quadratic formula to find how far his throw traveled horizontally to the nearest meter.

Explanation:

Problem 8

Step1: Set up the equation

We know \( h(t) = -16t^2 + 60t \) and we want to find \( t \) when \( h(t)=56 \). So we set up the equation:
\( -16t^2 + 60t = 56 \)
Rearrange it to standard quadratic form \( ax^2+bx + c = 0 \):
\( -16t^2 + 60t - 56 = 0 \)
We can simplify by dividing all terms by -4:
\( 4t^2 - 15t + 14 = 0 \)
Here, \( a = 4 \), \( b=- 15 \), \( c = 14 \)

Step2: Apply the quadratic formula

The quadratic formula is \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D=b^2-4ac \)
\( D=(-15)^2-4\times4\times14=225 - 224 = 1 \)
Then, find the roots:
\( t=\frac{15\pm\sqrt{1}}{2\times4}=\frac{15\pm1}{8} \)
For the plus sign: \( t_1=\frac{15 + 1}{8}=\frac{16}{8}=2 \)
For the minus sign: \( t_2=\frac{15-1}{8}=\frac{14}{8}=1.75 \)

Step1: Set up the equation

We know \( d = 0.05s^2+1.1s \) and \( d = 200 \). So we set up the equation:
\( 0.05s^2+1.1s=200 \)
Rearrange to standard quadratic form \( ax^2+bx + c = 0 \):
\( 0.05s^2+1.1s - 200=0 \)
Multiply all terms by 20 to eliminate decimals:
\( s^2 + 22s-4000 = 0 \)
Here, \( a = 1 \), \( b = 22 \), \( c=-4000 \)

Step2: Apply the quadratic formula

The quadratic formula is \( s=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D=b^2-4ac \)
\( D=(22)^2-4\times1\times(-4000)=484 + 16000=16484 \)
\( \sqrt{16484}\approx128.4 \) (we take the positive root since speed \( s>0 \))
Then, \( s=\frac{-22\pm128.4}{2} \)
We take the positive sign (because speed can't be negative):
\( s=\frac{-22 + 128.4}{2}=\frac{106.4}{2}=53.2 \)

Step1: Set up the equation

We want to find the horizontal distance \( x \) when the shot put hits the ground, so \( y = 0 \). The equation is \( y=-16x^2 + 368x \)
Set \( y = 0 \):
\( -16x^2+368x=0 \)
Factor out \( -16x \):
\( -16x(x - \frac{368}{16})=0 \)
Simplify \( \frac{368}{16} = 23 \)
So the equation is \( -16x(x - 23)=0 \)
We can also use the quadratic formula. For \( -16x^2+368x=0 \), \( a=-16 \), \( b = 368 \), \( c = 0 \)

Step2: Apply the quadratic formula

The quadratic formula is \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \)
Calculate the discriminant \( D=b^2-4ac=(368)^2-4\times(-16)\times0=368^2 \)
\( \sqrt{D}=368 \)
Then, \( x=\frac{-368\pm368}{2\times(-16)} \)
For the minus sign: \( x=\frac{-368 + 368}{-32}=0 \) (this is the starting point)
For the plus sign: \( x=\frac{-368-368}{-32}=\frac{-736}{-32}=23 \)

Answer:

The object will be at a height of 56 feet at \( t = 1.75 \) seconds and \( t=2 \) seconds.

Problem 9