Answer :
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.
