Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To solve the given problems, we'll use the equation for the height of the object, which is [tex]\( h = -16t^2 + 194t + 15 \)[/tex]. We'll solve for specific values of [tex]\( t \)[/tex] in two scenarios: when the height [tex]\( h \)[/tex] is 413 feet and when the object reaches the ground (i.e., the height [tex]\( h = 0 \)[/tex]).
### 1. Finding when the height will be 413 feet:
We start by setting the height [tex]\( h \)[/tex] to 413 feet and solve for [tex]\( t \)[/tex] in the equation:
[tex]\[ 413 = -16t^2 + 194t + 15 \][/tex]
Rearranging the equation to bring all terms to one side, we get:
[tex]\[ -16t^2 + 194t + 15 - 413 = 0 \][/tex]
[tex]\[ -16t^2 + 194t - 398 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 194 \)[/tex], and [tex]\( c = -398 \)[/tex]. Solving this quadratic equation using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ t = \frac{-194 \pm \sqrt{194^2 - 4(-16)(-398)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-194 \pm \sqrt{37636 - 25472}}{-32} \][/tex]
[tex]\[ t = \frac{-194 \pm \sqrt{12164}}{-32} \][/tex]
[tex]\[ t = \frac{-194 \pm 110.3126}{-32} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = \frac{-194 + 110.3126}{-32} = \frac{-83.6874}{-32} = 2.615 \text{ seconds} \][/tex]
[tex]\[ t_2 = \frac{-194 - 110.3126}{-32} = \frac{-304.3126}{-32} = 9.510 \text{ seconds} \][/tex]
So, the height will be 413 feet at approximately [tex]\( 2.615 \)[/tex] seconds and [tex]\( 9.510 \)[/tex] seconds.
### 2. Finding when the object will reach the ground:
To determine when the object reaches the ground, we set the height [tex]\( h \)[/tex] to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = -16t^2 + 194t + 15 \][/tex]
This is again a quadratic equation. Solving for [tex]\( t \)[/tex] using the quadratic formula:
[tex]\[ t = \frac{-194 \pm \sqrt{194^2 - 4(-16)(15)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-194 \pm \sqrt{37636 + 960}}{-32} \][/tex]
[tex]\[ t = \frac{-194 \pm \sqrt{38596}}{-32} \][/tex]
[tex]\[ t = \frac{-194 \pm 196.4533}{-32} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = \frac{-194 + 196.4533}{-32} = \frac{2.4533}{-32} = -0.077 \text{ seconds} \][/tex]
[tex]\[ t_2 = \frac{-194 - 196.4533}{-32} = \frac{-390.4533}{-32} = 12.202 \text{ seconds} \][/tex]
Therefore, because time cannot be negative, the object will reach the ground at approximately [tex]\( 12.202 \)[/tex] seconds.
### Summary:
- The object will be 413 feet high at approximately [tex]\( 2.615 \)[/tex] seconds and [tex]\( 9.510 \)[/tex] seconds.
- The object will reach the ground at approximately [tex]\( 12.202 \)[/tex] seconds.
### 1. Finding when the height will be 413 feet:
We start by setting the height [tex]\( h \)[/tex] to 413 feet and solve for [tex]\( t \)[/tex] in the equation:
[tex]\[ 413 = -16t^2 + 194t + 15 \][/tex]
Rearranging the equation to bring all terms to one side, we get:
[tex]\[ -16t^2 + 194t + 15 - 413 = 0 \][/tex]
[tex]\[ -16t^2 + 194t - 398 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 194 \)[/tex], and [tex]\( c = -398 \)[/tex]. Solving this quadratic equation using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ t = \frac{-194 \pm \sqrt{194^2 - 4(-16)(-398)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-194 \pm \sqrt{37636 - 25472}}{-32} \][/tex]
[tex]\[ t = \frac{-194 \pm \sqrt{12164}}{-32} \][/tex]
[tex]\[ t = \frac{-194 \pm 110.3126}{-32} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = \frac{-194 + 110.3126}{-32} = \frac{-83.6874}{-32} = 2.615 \text{ seconds} \][/tex]
[tex]\[ t_2 = \frac{-194 - 110.3126}{-32} = \frac{-304.3126}{-32} = 9.510 \text{ seconds} \][/tex]
So, the height will be 413 feet at approximately [tex]\( 2.615 \)[/tex] seconds and [tex]\( 9.510 \)[/tex] seconds.
### 2. Finding when the object will reach the ground:
To determine when the object reaches the ground, we set the height [tex]\( h \)[/tex] to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = -16t^2 + 194t + 15 \][/tex]
This is again a quadratic equation. Solving for [tex]\( t \)[/tex] using the quadratic formula:
[tex]\[ t = \frac{-194 \pm \sqrt{194^2 - 4(-16)(15)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-194 \pm \sqrt{37636 + 960}}{-32} \][/tex]
[tex]\[ t = \frac{-194 \pm \sqrt{38596}}{-32} \][/tex]
[tex]\[ t = \frac{-194 \pm 196.4533}{-32} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = \frac{-194 + 196.4533}{-32} = \frac{2.4533}{-32} = -0.077 \text{ seconds} \][/tex]
[tex]\[ t_2 = \frac{-194 - 196.4533}{-32} = \frac{-390.4533}{-32} = 12.202 \text{ seconds} \][/tex]
Therefore, because time cannot be negative, the object will reach the ground at approximately [tex]\( 12.202 \)[/tex] seconds.
### Summary:
- The object will be 413 feet high at approximately [tex]\( 2.615 \)[/tex] seconds and [tex]\( 9.510 \)[/tex] seconds.
- The object will reach the ground at approximately [tex]\( 12.202 \)[/tex] seconds.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
