# Projectile on an inclined plane

Say you have the classic problem of a projectile going at 10m/s at 15 degrees off a plane already inclined at 30 degrees (so the projectile's angle with respect to the earth is 45 degrees), and you are asked for the time the projectile was in the air. My question is not so simply how to work it, but rather why one way of working doesn't work.

Imagine one were to tilt their head 30 degrees, so that the "hill" is no longer inclined but is now flat earth. Regardless of how you tilt your head, earth's 9.8 m/s/s acceleration will be pointed 30 degrees skew from perpendicular to the hill. However it stands to reason that you can "convert" earth's acceleration to match the hill's normal vector.

In the new perspective, gravity's "downward force" (perpendicular to the hill) is now 8.49 m/s/s, and now there's a horizontal acceleration pulling left 4.9 m/s/s. You also have to convert your initial velocity of 10 m/s with a 15 degree incline giving a "downward" velocity of 2.59 m/s.

Now you know acceleration (8.49 m/s/s), velocity (2.59 m/s), and your change in "height"/distance from the hill's surface (0 m). Since you need time in the air, you can use the kinematic equation:

$$\Delta x = V_iT + \frac 1 2 aT^2$$

Solving this equation yields the wrong answer. I don't understand why this approach doesn't work considering that all I'm trying to find is air-time, not displacement. What I'd expect to be wrong is the conversion of gravitational acceleration from the tilt, but I can't find the problem there. I'm not asking how to best work the problem, I'm asking why this method doesn't work.

What you're trying to do is called a transformation of the coordinate system (specifically a rotation). In the new coordinate system, shown as the $xy$-axis, the line of incline becomes the $x$-axis and the Normal to that, the $y$-axis. We choose as origin $(0,0)$ of the new coordinate system the start of the incline.

As a result, the Earth's acceleration vector $\vec{g}$ now has to be decomposed into the components:

$$g_y=g\cos\phi$$ $$g_x=g\sin\phi$$

Similarly the initial velocity vector $\vec{v}_0$ has to be decomposed:

$$v_{0,y}=v_0\sin(\theta_i-\phi)$$ $$v_{0,x}=v_0\cos(\theta_i-\phi)$$

The $y$-component of velocity can now be written as:

$$v_y=v_{0,y}-g_yt=v_0\sin(\theta_i-\phi)-g(\cos\phi)t\tag{1}$$

And the $x$-component:

$$v_x=v_0\cos(\theta_i-\phi)-g(\sin\phi)t\tag{2}$$

Use $(1)$ to find the time where $v_y$ becomes $0$. Due to symmetry this is half the flight time.

Equation $(2)$ doesn't yet give you the displacenent $x$: it needs to be integrated to time $t$ once to obtain $x(t)$, Then insert the total flight time into that expression, to get $d$.

So your idea is workable but whether such a derivation is really simpler than doing it in the original vertical-horizontal system remains to be seen.

Otherwise, you can use $$T=\dfrac{2u\sin \left( \theta -\phi \right) }{g\cos \phi }$$ $$=\dfrac{2\cdot 10\sin \left( 45-30\right)^{\circ } }{9.8\cos 30^{\circ }}$$