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

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:

$x -y -1.7675=0$