Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To solve these problems, we will use the equation for height, [tex]\( h = -16t^2 + 136t + 18 \)[/tex], where [tex]\( t \)[/tex] is the time in seconds.
### Part 1: When will the height be 252 feet?
We need to determine the value(s) of [tex]\( t \)[/tex] when [tex]\( h = 252 \)[/tex]:
[tex]\[ 252 = -16t^2 + 136t + 18 \][/tex]
Rearranging this equation, we get:
[tex]\[ -16t^2 + 136t + 18 - 252 = 0 \][/tex]
[tex]\[ -16t^2 + 136t - 234 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
[tex]\[ a = -16 \][/tex]
[tex]\[ b = 136 \][/tex]
[tex]\[ c = -234 \][/tex]
We solve this using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-136 \pm \sqrt{136^2 - 4(-16)(-234)}}{2(-16)} \][/tex]
Solving the discriminant part [tex]\( b^2 - 4ac \)[/tex] first:
[tex]\[ 136^2 = 18496 \][/tex]
[tex]\[ -4 \cdot (-16) \cdot (-234) = 14976 \][/tex]
[tex]\[ b^2 - 4ac = 18496 - 14976 = 3520 \][/tex]
Thus, the equation for [tex]\( t \)[/tex] becomes:
[tex]\[ t = \frac{-136 \pm \sqrt{3520}}{-32} \][/tex]
Simplifying:
[tex]\[ \sqrt{3520} = \sqrt{16 \cdot 220} = 4\sqrt{220} \][/tex]
So,
[tex]\[ t = \frac{-136 \pm 4\sqrt{220}}{-32} \][/tex]
[tex]\[ t = \frac{-136 + 4\sqrt{220}}{-32}, \quad t = \frac{-136 - 4\sqrt{220}}{-32} \][/tex]
Further simplifying:
[tex]\[ t = \frac{-136 + \sqrt{3520}}{-32} = \frac{-136}{-32} + \frac{4\sqrt{220}}{-32} \][/tex]
[tex]\[ t = \frac{136}{32} - \frac{\sqrt{220}}{8} \][/tex]
[tex]\[ t = \frac{255}{32} - \frac{\sqrt{220}}{8} \][/tex]
The solutions [tex]\( t \)[/tex] are:
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{55}}{4}, \quad t = \frac{17}{4} + \frac{\sqrt{55}}{4} \][/tex]
Therefore, the height will be 252 feet at:
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{55}}{4} \][/tex]
[tex]\[ t = \frac{17}{4} + \frac{\sqrt{55}}{4} \][/tex]
### Part 2: When will the object reach the ground?
We need to determine the value(s) of [tex]\( t \)[/tex] when [tex]\( h = 0 \)[/tex]:
[tex]\[ 0 = -16t^2 + 136t + 18 \][/tex]
Solving the quadratic equation:
[tex]\[ -16t^2 + 136t + 18 = 0 \][/tex]
Using the same quadratic formula:
[tex]\[ t = \frac{-136 \pm \sqrt{136^2 - 4(-16)(18)}}{2(-16)} \][/tex]
Solving the discriminant:
[tex]\[ 136^2 = 18496 \][/tex]
[tex]\[ -4 \cdot (-16) \cdot 18 = 1152 \][/tex]
[tex]\[ b^2 - 4ac = 18496 + 1152 = 19648 \][/tex]
Thus:
[tex]\[ t = \frac{-136 \pm \sqrt{19648}}{-32} \][/tex]
Simplifying:
[tex]\[ \sqrt{19648} = \sqrt{64 \cdot 307} = 8\sqrt{307} \][/tex]
So,
[tex]\[ t = \frac{-136 \pm 8\sqrt{307}}{-32} \][/tex]
[tex]\[ t = \frac{-136}{-32} + \frac{8\sqrt{307}}{-32} \][/tex]
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{307}}{4} \][/tex]
The solutions [tex]\( t \)[/tex] are:
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{307}}{4}, \quad t = \frac{17}{4} + \frac{\sqrt{307}}{4} \][/tex]
Therefore, the object will reach the ground at:
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{307}}{4} \][/tex]
[tex]\[ t = \frac{17}{4} + \frac{\sqrt{307}}{4} \][/tex]
These results give us the times at which the height is 252 feet and when the object reaches the ground.
### Part 1: When will the height be 252 feet?
We need to determine the value(s) of [tex]\( t \)[/tex] when [tex]\( h = 252 \)[/tex]:
[tex]\[ 252 = -16t^2 + 136t + 18 \][/tex]
Rearranging this equation, we get:
[tex]\[ -16t^2 + 136t + 18 - 252 = 0 \][/tex]
[tex]\[ -16t^2 + 136t - 234 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
[tex]\[ a = -16 \][/tex]
[tex]\[ b = 136 \][/tex]
[tex]\[ c = -234 \][/tex]
We solve this using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-136 \pm \sqrt{136^2 - 4(-16)(-234)}}{2(-16)} \][/tex]
Solving the discriminant part [tex]\( b^2 - 4ac \)[/tex] first:
[tex]\[ 136^2 = 18496 \][/tex]
[tex]\[ -4 \cdot (-16) \cdot (-234) = 14976 \][/tex]
[tex]\[ b^2 - 4ac = 18496 - 14976 = 3520 \][/tex]
Thus, the equation for [tex]\( t \)[/tex] becomes:
[tex]\[ t = \frac{-136 \pm \sqrt{3520}}{-32} \][/tex]
Simplifying:
[tex]\[ \sqrt{3520} = \sqrt{16 \cdot 220} = 4\sqrt{220} \][/tex]
So,
[tex]\[ t = \frac{-136 \pm 4\sqrt{220}}{-32} \][/tex]
[tex]\[ t = \frac{-136 + 4\sqrt{220}}{-32}, \quad t = \frac{-136 - 4\sqrt{220}}{-32} \][/tex]
Further simplifying:
[tex]\[ t = \frac{-136 + \sqrt{3520}}{-32} = \frac{-136}{-32} + \frac{4\sqrt{220}}{-32} \][/tex]
[tex]\[ t = \frac{136}{32} - \frac{\sqrt{220}}{8} \][/tex]
[tex]\[ t = \frac{255}{32} - \frac{\sqrt{220}}{8} \][/tex]
The solutions [tex]\( t \)[/tex] are:
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{55}}{4}, \quad t = \frac{17}{4} + \frac{\sqrt{55}}{4} \][/tex]
Therefore, the height will be 252 feet at:
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{55}}{4} \][/tex]
[tex]\[ t = \frac{17}{4} + \frac{\sqrt{55}}{4} \][/tex]
### Part 2: When will the object reach the ground?
We need to determine the value(s) of [tex]\( t \)[/tex] when [tex]\( h = 0 \)[/tex]:
[tex]\[ 0 = -16t^2 + 136t + 18 \][/tex]
Solving the quadratic equation:
[tex]\[ -16t^2 + 136t + 18 = 0 \][/tex]
Using the same quadratic formula:
[tex]\[ t = \frac{-136 \pm \sqrt{136^2 - 4(-16)(18)}}{2(-16)} \][/tex]
Solving the discriminant:
[tex]\[ 136^2 = 18496 \][/tex]
[tex]\[ -4 \cdot (-16) \cdot 18 = 1152 \][/tex]
[tex]\[ b^2 - 4ac = 18496 + 1152 = 19648 \][/tex]
Thus:
[tex]\[ t = \frac{-136 \pm \sqrt{19648}}{-32} \][/tex]
Simplifying:
[tex]\[ \sqrt{19648} = \sqrt{64 \cdot 307} = 8\sqrt{307} \][/tex]
So,
[tex]\[ t = \frac{-136 \pm 8\sqrt{307}}{-32} \][/tex]
[tex]\[ t = \frac{-136}{-32} + \frac{8\sqrt{307}}{-32} \][/tex]
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{307}}{4} \][/tex]
The solutions [tex]\( t \)[/tex] are:
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{307}}{4}, \quad t = \frac{17}{4} + \frac{\sqrt{307}}{4} \][/tex]
Therefore, the object will reach the ground at:
[tex]\[ t = \frac{17}{4} - \frac{\sqrt{307}}{4} \][/tex]
[tex]\[ t = \frac{17}{4} + \frac{\sqrt{307}}{4} \][/tex]
These results give us the times at which the height is 252 feet and when the object reaches the ground.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.