Angle psr measures 99°. what is the measure of psq in degrees?
1 answer:
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So, notice, the focus point is at -7, 5, and the directrix is at y = -11.
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.
![\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-7\\ k=-3\\ p=8 \end{cases}\implies 4(8)[y-(-3)]=[x-(-7)]^2 \\\\\\ 32(y+3)=(x+7)^2\implies y+3=\cfrac{1}{32}(x+7)^2 \\\\\\ y=\cfrac{1}{32}(x+7)^2-3](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A4p%28x-%20h%29%3D%28y-%20k%29%5E2%0A%5C%5C%5C%5C%0A%5Cboxed%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%0A%5Cend%7Barray%7D%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%0A%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ah%3D-7%5C%5C%0Ak%3D-3%5C%5C%0Ap%3D8%0A%5Cend%7Bcases%7D%5Cimplies%204%288%29%5By-%28-3%29%5D%3D%5Bx-%28-7%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A32%28y%2B3%29%3D%28x%2B7%29%5E2%5Cimplies%20y%2B3%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2%0A%5C%5C%5C%5C%5C%5C%0Ay%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2-3)
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.
Answer:
We assume that the orange is dropped at t = 0s.
Once the orange is on the air, the only force acting on it is the gravitational force, then the acceleration of the orange is the gravitational acceleration.
A(t) = -32.17 ft/s^2
Where the negative sign is because this acceleration points downwards.
For the velocity equation, we need to integrate over time, we will get:
V(t) = (-32.17 ft/s^2)*t + V0
Where V0 is the initial vertical velocity of the orange, because the orange is accidentally dropped, this initial velocity is equal to zero.
V(t) = (-32.17 ft/s^2)*t
For the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-32.17 ft/s^2)*t^2 + P0
Where P0 is the initial height of the orange, we know that it is 40ft, then the position equation is:
P(t) = (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft
Now that we know the equation, we can graph it. (you can see the graph below)
Now we also want to find at what time does the orange hit the water.
This happens when:
P(t) = 0 ft = (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft
We just need to solve that equation for t.
0 ft = (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft
(1/2)*(32.17 ft/s^2)*t^2 = 40 ft
t^2 = (40ft)/( (1/2)*(32.17 ft/s^2))
t = √( (40ft)/( (1/2)*(32.17 ft/s^2)) ) = 1.58 s
The orange hits the water 1.58 seconds after it is dropped.
