Respuesta :
The horizontal function is x(t) = 8·t
The vertical function is y(t) = -16·t² + 100
A) The answer is: it takes her 2.62 time units to reach the ground.
The time taken to move horizontally is equal to the time taken to move vertically from the cliff to the ground, therefore, we can set:
8·t = -16·t² + 100
and solve for t:
16·t² + 8·t - 100 = 0
4(4·t² + 2·t - 25) = 0
4·t² + 2·t - 25 = 0
[tex]t_{1,2} = \frac{- \frac{b}{2} \pm \sqrt{(\frac{b}{2})^{2} - ac } }{a} [/tex]
[tex]t_{1,2} = \frac{- \frac{2}{2} \pm \sqrt{(\frac{2}{2})^{2} - (4)(-25) } }{4}[/tex]
t₁ = (-1 + √101) / 4 = 2.62
t₂ = (-1 - √101) / 4 = - 2.76
We cannot accept negative times, therefore the solution is t = 2.62 time-units.
B) The answer is: she lands 20.96 distance-units from the cliff.
In order to find the distance travelled, substitute the time taken found in point A into the equation of the horizontal movement (using the vertical one you would calculate the height of the cliff):
x(t) = 8·t
x(2.62) = 8 · 2.62 = 20.96
Hence, Lukalu lands 20.96 distance-units from the cliff.
The vertical function is y(t) = -16·t² + 100
A) The answer is: it takes her 2.62 time units to reach the ground.
The time taken to move horizontally is equal to the time taken to move vertically from the cliff to the ground, therefore, we can set:
8·t = -16·t² + 100
and solve for t:
16·t² + 8·t - 100 = 0
4(4·t² + 2·t - 25) = 0
4·t² + 2·t - 25 = 0
[tex]t_{1,2} = \frac{- \frac{b}{2} \pm \sqrt{(\frac{b}{2})^{2} - ac } }{a} [/tex]
[tex]t_{1,2} = \frac{- \frac{2}{2} \pm \sqrt{(\frac{2}{2})^{2} - (4)(-25) } }{4}[/tex]
t₁ = (-1 + √101) / 4 = 2.62
t₂ = (-1 - √101) / 4 = - 2.76
We cannot accept negative times, therefore the solution is t = 2.62 time-units.
B) The answer is: she lands 20.96 distance-units from the cliff.
In order to find the distance travelled, substitute the time taken found in point A into the equation of the horizontal movement (using the vertical one you would calculate the height of the cliff):
x(t) = 8·t
x(2.62) = 8 · 2.62 = 20.96
Hence, Lukalu lands 20.96 distance-units from the cliff.
By solving quadratic equation we got that it take 2.62 unit time her to reach the ground and she is 20.96 unit distance away from the cliff is when she lands.
What is quadratic equation ?
Any equation of the form [tex]ax^2+bx+c=0[/tex]where x is variable and a, b, and c are any real numbers where a ≠ 0 is called quadratic equation .
Here given that Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of time are x(t)=8t and y(t)=-16[tex]t^2[/tex]+100
Now the time taken to move horizontally is equal to the time taken to move vertically from the cliff to the ground
[tex]8t=-16t^2+100\\\\16t^2+8t-100=0\\\\[/tex]
Using quadratic formula
[tex]t=\frac{-8\pm \sqrt(8^2+4\times\times\100)}{2\times16}[/tex]
t=2.62,-2.76
t can't be negative hence t= 2.62
Now distance travelled =x(2.62)= 8(2.62)=20.96
By solving quadratic equation we got that it take 2.62 unit time her to reach the ground and she is 20.96 unit distance away from the cliff is when she lands.
To learn more about quadratic equations visit : https://brainly.com/question/1214333
