Show that for a projectile the angle between the velocity and the *x*-axis as a function of time is given by

`theta(t) =tan^(-1) ((v_(0y) - "gt")/v_(o x))`

#### Solution 1

Let `v_(ox)` and `v_(oy)` respectively be the initial components of the velocity of the projectile along horizontal (*x*) and vertical (*y*) directions.

Let `v_x` and `v_y` respectively be the horizontal and vertical components of velocity at a point P.

Time taken by the projectile to reach point P = *t*

Applying the first equation of motion along the vertical and horizontal directions, we get:

`v_y = v_(oy) = "gt"`

And `v_x =v_(ox)`

`:. tan theta = v_y/v_x= (v_(oy) - "gt")/v_(ox)`

`theta = tan^(-1)((v_(oy) - "gt")/(v_(o x)))`

#### Solution 2

Let the projectile be fired at an angle `theta` with x-axis.

As `theta` depend on t, `theta(t)` at any instance

`tan theta(t) = v_y/v_x =(v_(oy) - "gt")/(v_(o x))`

(Since `v_y = v_(oy) - "gt" and v_x = v_(o x)`)

`=> theta(t) = tan^(-1)((v_o - "gt")/v_(o x))`