Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
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.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
