Answer :
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.
