Both of the equations listed here are helpful but the best equation to use is combining the two equations;
x = vt cos θ
y = vt sin θ - ½ g t²
From the first one, get t on its own and sub it into the second one. You don't need to take into account what the time is for your problem. The math would get a bit ugly on this, but after simplifying you should get;
y = x tanθ - gx²(1+ tan² θ)
So if you need to hit a target x metres away and y metres high, you can work out the angle needed for the velocity you launch it at. Or if you launch it at an angle theta, velocity v and you want to know how far away the projectile is when it is one meter high...
Mathematically for (a), you need to prove that each side of the quadrilateral is equal in magnitude and that opposing sides of the quadrilateral are parallel and perpendicular to the other sides.
Length is easy
length = √[(x2-x1)² + (y2-y1)²]
So the length from (-3,1) to (-1,4) is √[(-1--3)² + (4-1)²] = √[4 + 9] = √13
And the length from (-3,1) to (0,-1) is √[(0--3)² + (-1-1)²] = √[9 + 4] = √13
To prove they are perpendicular to each other just find the gradients of each side and two of the lines will have gradient m whilst the other two sides will have gradient -1/m
The gradient can be found using (y2 - y1)/(x2 - x1)
So the gradient from (-3,1) to (-1,4) is => (4-1)/(-1-1) = -3/2
And the gradient from (-3,1) to (0,-1) is => (-1-1)/(0--3) = 2/3
So just do this for the other two sides, proving their lengths are root 13, and their gradients are m, -1/m, m and -1/m.
(b) is similar. Work out the equations of the diagonals using y = mx + c, filling in the values of y and x for each pair of diagnol coordinates. Then show that the gradient, m, for one diagonal is perpendicular to the other diagnol with gradient -1/m.
(c) is just working out the length of the diagonals using the formulae i gave you in part (a). Simple enough. The values of these lines should come out as √26 since this is what pythagorus theorem suggests.
Using pythagorus theorem => a² = b² + c²
a² = (√13)² + (√13)²
a² = 13 + 13 = 26
a = √26
Length = √[(2 - 1)² + (2 - -3)²] = √[1 + 25] = √26
Above is an image of what i mean. The green shows random trajectory values for theta greater than 45. The blue line is the path for theta equals 45. The orange line is the path chosen by theta equal to 90 degrees (i.e. straight up and down the y-axis)
I want to find the equation of the red path. If i take any point on this red curve or the area below it, I could find a theta for the trajectory to reach this point. If I use a point exactly on this line then I'm only going to get one repeated solution for theta. And if I chose any point above the red line, then there are no solutions for theta since the magnitude of v is not great enough.
Hope that makes it clearer.
I know how to find the maximum range and the maximum height. I also know the trajectory equation which will tell me who to find the displacement from the origin.
What I want to know, and it's hard to put, is the equation for the maximum point of displacements for a fixed velocity v, but a variant angle.
Imagine you draw out the axis, and set u= 100. Then for theta=1, you draw the path. Then for theta=2,...every single value of theta from 0 to 90. If done by a computer, the final image produced would look like a -x^2 graph with the area shaded. I want to know the equation of that curve given i.e. the maximum displacement from the origin.
If an object is projected at a fixed speed of u at a variable angle theta, how would one find out the equation of the curve that shows the furthest point the trajectory can be?
So if the theta= 90, then the body goes to it's highest possible point and is also where the curve would cut the y-axis. When theta = 45, the x value is v2/g (the maximum range) and shows where the curve cuts the x axis. So what is the equation of the curve that shows the maximum position the trajectory can travel to for any value of theta.