Question

Show that for a projectile, the angle between velocity and the X-axis as a function of time is given by
$\theta \left(t\right)={\tan }^{-1}\frac{v_{0y}-gt}{v_{0}x}$

Answer 1

For a projectile motion along $x$ axis is given as,

$x=v_{0x}t$

For a projectile motion along $x$ axis is given as,

$y=v_{0y}t-\frac{1}{2}g{t}^2$

Slope of the two gives the angle between the velocity component of $x$ axis and $y$ axis.

$\frac{dy}{dt}=v_{0y}-gt$

$\frac{dx}{dt}=v_{0x}$

$\frac{dy}{dx}=\frac{v_{0y}-gt}{v_{0x}}$

$\theta \left(t\right)={\tan }^{-1}\left(\frac{v_{0y}-gt}{v_{0x}}\right)$