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})](https://tex.z-dn.net/?f=%28%5Cfrac%7B1.25%7D%7B%5Csqrt2%7D%2C%5Cfrac%7B-1.25%7D%7B%5Csqrt2%7D%29)
Taking O as center and Coordinates of O as (0,0) and Coordinates of M as (
) 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}{\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](https://tex.z-dn.net/?f=%20%5Cfrac%7B1.25%7D%7B%5Csqrt%202%7D-%20%5Cfrac%7B-1.25%7D%7B%5Csqrt%202%7D%2Bk%3D0%5C%5C%5C%5C%202%20%5Ctimes%5Cfrac%7B1.25%7D%7B%5Csqrt%202%7D%2Bk%3D0%5C%5C%5C%5C1.7675%2Bk%3D0%5C%5C%5C%5C%20k%3D-1.7675)
So, equation of initial path of stone will be:
![x -y -1.7675=0](https://tex.z-dn.net/?f=%20x%20-y%20-1.7675%3D0)