newtonian-gravityvectors
So I was writing a lab report for Physics I, where I was describing projectile motion. I had written
Since gravity on Earth does not have a horizontal force, the object
will not lose horizontal velocity (velocity in the x direction).
Then this made me think, is there any gravitational force found in nature that has both vertical and horizontal components?
The formula you have written is correct; but they are functions of time. Hence, by inserting the particular instant , say $t$ on the function ,you get the instantaneous components of velocity. Then using phythagoras theorem you will get the total instantaneous velocity. Taking your example, at time $T$ s , the X-comp. is $30$ unit and Y-comp. is $(20 - 9.81T) $ unit . So,velocity at time $T$ is $\sqrt{(30)^2 + (20 - 9.81T)^2}$ unit
I find this type of question is always easier if you draw a diagram. The drag force $F$ acts in the opposite direction to the velocity so it looks like:
and the components of the drag force are:
$$\begin{align} F_x &= F \cos\theta \\ F_y &= F \sin\theta \end{align}$$
$\cos\theta$ is $v_x/v$ and $v = \sqrt{v_x^2 + v_y^2}$ so we get:
$$ F_x = F \frac{v_x}{\sqrt{v_x^2 + v_y^2}} $$
Since $F = -Cv^2$ we get:
$$ F_x = -Cv^2 \frac{v_x}{\sqrt{v_x^2 + v_y^2}} $$
and because $v = \sqrt{v_x^2 + v_y^2}$ this becomes:
$$\begin{align} F_x &= -C \left(v_x^2 + v_y^2\right) \frac{v_x}{\sqrt{v_x^2 + v_y^2}} \\ &= -C v_x \sqrt{v_x^2 + v_y^2} \end{align}$$
And likewise for $F_y$.
Best Answer
If you think about it, "vertical" is generally defined as the direction in which things fall - that is, parallel to the local gravitational field. So by that definition of "vertical," no, it is not possible. But if you were to use a different definition of "vertical" and "horizontal," sure, it might be possible.