A tennis player hits a ball at point o which is at a height of 2 meters above the ground and at a horizontal distance 4metres from the net. The initial velocity being at a direction of 45 above the horizontal. If the ball just clears the net which is 1 meter high and taking acceleration due to gravity as 10, show that the equation of the path of the ball is 16y=6x−5xsquared

Step 1: Define the coordinate system and the general equation of the path.
Let the point where the ball is hit be the origin $(0,0)$ of our coordinate system.
The initial height of the ball above the ground is $h_0 = 2 \text{ m}$.
The initial velocity is $u$ at an angle $\theta = 45^\circ$ above the horizontal.
The acceleration due to gravity is $g = 10 \text{ m/s}^2$.

The general equation for the trajectory of a projectile launched from an initial height $h_0$ (relative to the ground) is:

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