Find solutions to your problems with the help of IDNLearn.com's knowledgeable users. Our platform offers reliable and detailed answers, ensuring you have the information you need.
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 appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.
