Question

Ancient slingshots were made form straps of leather that cradled a rock until it was released. One would spin the slingshot in a circle, and the initial path of the released rock would be a straight line tangent to the circle at the point of release. The rock will travel the greatest distance if it is release when the angle between the normal to the path and the horizontal is -45 degrees. The center of the circular path is the origin and the radius of the circle measure 1.25 feet. Write and equation of the initial path of the rock in standard form.

Answer 1

We separate equation by looking at x and y projection separately. 
Let us look at the x projection first.
Motion starts with initial velocity $v_0cos(45)$ because there is no force acting along the x-axis we don't have any acceleration along the x-axis.  Our equation would be:
$x=v_0cos(45)\cdot t$
Now our y equation will be a little bit more complicated. There is gravity working against our projectile and we also have to take into effect the hight from which the projectile is launched.
With all that in mind let us write the equation for y projection:
$y=h+v_0sin(45)\cdot t-g \frac{t^2}{2}$
The term $-g \frac{t^2}{2}$ represents gravity and as you can see it is negative, which means gravity is pulling our projectile down. 
There is one more thing we can do. We can express initial velocity $v_0$ in relation to circular motion used to launch the projectile.
$v_0=w\cdot r$
Where r is the length of the letter used to launch the projectile, and $w$ is angular velocity.  
If we combine our x and y equations we can get the trajectory of our projectile.
$y=h+x- \frac{gx^2}{2\cdot (wr)^2cos^2(45)}$
This is the equation of a parabola. If you follow this link(https://www.desmos.com/calculator/vogcuygjhs) it will take to an interactive graph where you can see how trajectory looks like depending on some of the paramaters.

Answer 2

Solution:

Taking ON as horizontal axis and OP as vertical axis.

Suppose from the , point M stone is released and it travels the greatest distance.  

∠ MON= -45°→→ angle lies in fourth quadrant.

Where, MN represents the path of rock, which is a straight line.

Taking the Horizontal and vertical component of OM,to get the coordinate of point M

The coordinate of point M will be [1.25 cos (-45°), 1.25 sin(-45°)]=$(\frac{1.25}{\sqrt2},\frac{-1.25}{\sqrt2})$

Taking O as center and Coordinates of O as (0,0) and Coordinates of M as ($(\frac{1.25}{\sqrt2},\frac{-1.25}{\sqrt2})$) we get the equation of line OM as

y= tan (315°)× x

As general form of equation of line passing through the origin will be,

y= m x, where m is angle made by line with positive direction of x axis.

m= (360-45)°= 315°

→→y=-x

→→  y +  x=0

equation of line perpendicular to OM will be

→→ x -  y + k=0

Since it passes through $(\frac{1.25}{\sqrt2},\frac{-1.25}{\sqrt2})$.

→→$\frac{1.25}{\sqrt 2}- \frac{-1.25}{\sqrt 2}+k=0\\\\ 2 \times\frac{1.25}{\sqrt 2}+k=0\\\\1.7675+k=0\\\\ k=-1.7675$

So, equation of initial path of stone will be:

$x -y -1.7675=0$