Derive Equations 8.3 for the acceleration of a projectile subject to drag.

Consider a ball of mass$m$ in a projectile motion with initial velocity$v$. Let the trajectory of the ball makes an angle $\theta$ with the $x$ axis.

Let's consider the $x$-component of the velocity as follow:

$v_x=v\cos \theta ⟹\cos \theta =\frac{v_x}{v}=\frac{v_x}{\sqrt{v_x^2+v_y^2}}$

Then, the$y$-component of the velocity is same as above, and it is

$v_y=v\sin \theta ⟹\sin \theta =\frac{v_y}{v}=\frac{v_y}{\sqrt{v_x^2+v_y^2}}$

The magnitude of the drag force is given by

$F_D=\frac{\rho v^{2}CA}{2}$

where $\rho$indicates the density of the air, $v$remains the speed of the ball, $C$shows the drag coefficient and$A$means the projected area of the ball.

The$x$component of the drag force is given by

${\left(F_D\right)}_x=-\frac{\rho v^{2}CA}{2}\cos \theta =-\frac{\rho v^{2}CA{v}_x}{2\sqrt{v_x^2+v_y^2}}=-\frac{\rho CA{v}_x\sqrt{v_x^2+v_y^2}}{2}$

To find the final expression, we are using the given equation.

$v^2=v_x^2+v_{y^{\text{'}}}^2$

Similarly the $y$component of the drag force is given as follow

${\left(F_D\right)}_y=-\frac{\rho v^{2}CA}{2}\sin \theta =-\frac{\rho v^{2}CA{v}_y}{2\sqrt{v_x^2+v_y^2}}=-\frac{\rho CA{v}_y\sqrt{v_x^2+v_y^2}}{2}$

Let's apply the Newton 's second law of motion along the$x$direction,

$m{a}_x={\left(F_D\right)}_x$

$m{a}_x=-\frac{\rho CA{v}_x\sqrt{v_x^2+v_y^2}}{2}$

$a_x=-\frac{\rho CA{v}_x\sqrt{v_x^2+v_y^2}}{2m}$

As same as the step 5, we are applying Newton's second law of motion along $y$direction,

$m{a}_y=-mg+{\left(F_D\right)}_y$

$m{a}_y=-mg-\frac{\rho CA{v}_y\sqrt{v_x^2+v_y^2}}{2}$

$a_y=-g-\frac{\rho CA{v}_y\sqrt{v_x^2+v_y^2}}{2m}$

Newton's second law of motion along $x$direction $a_x=-\frac{\rho CA{v}_x\sqrt{v_x^2+v_y^2}}{2m}$.

Newton' s second law of motion along $y$direction,

$a_y=-g-\frac{\rho CA{v}_y\sqrt{v_x^2+v_y^2}}{2m}$.