Time of flight for projectile on inclined plane

Question

A particle $P$ of mass $m$ lies on a plane inclined at an angle $\alpha$ to the direction vector $\mathbf{i}$. At $t=0$ the particle is projected from the origin of the coordinate system with speed $U$ at an angle $\theta$ to the plane. The particle is in a gravitational field $-g\,\mathbf{j}.$

Show that the particle hits the plane at time $t = \frac{2U\sin(\theta)}{g\cos(\alpha)}$, by considering the motion perpendicular to the plane.

I'm really stuck here. I have resolved the weight into perpendicular and parallel components, so that the perpendicular component of the weight is $-mg\cos(\alpha)\,\mathbf{j}$. Then I used $N2L$ which gives the acceleration as $-g\cos(\alpha)\,\mathbf{j}$. But this is where I get stuck.

I'm unsure how to incorporate the velocity in to this?

Answer 1

Hint:

The equations of motion are ( for $x$ horizontal axis and $y$ vertical axis): $$ x=\left[u\cos(\alpha+\theta)\right]t $$ $$ y=\left[u\sin(\alpha+\theta)\right]t-\frac{1}{2}gt^2 $$ where $u\cos(\alpha+\theta)$ and $u\sin(\alpha+\theta)$ are the components of the speed $v$ in the $x$ and $y$ direction.

And the equation of the plane is $$ y=x\tan \alpha $$

So the particle is on the plane when: $$ \left[u\sin(\alpha+\theta)\right]t-\frac{1}{2}gt^2=\tan \alpha \left[u\cos(\alpha+\theta)\right]t $$

solve for $t$ and you find the result.

Answer 2

The component perpendicular to the plane is not $-mg\cos(\alpha)\,\mathbf{j}$, because $\mathbf{j}$ is not perpendicular to the plane according to the way your question was formulated (unless $\alpha = 0$).

But that seems to be just a notational inconsistency. If you remove the direction vector $\mathbf{j}$ from that expression, you're left with the magnitude $-mg\cos(\alpha)$, which is the correct magnitude of acceleration perpendicular to the plane.

This exercise is based on the idea that in the ordinary (level plane) projectile motion analysis, the time spent in the air is entirely determined by the vertical component of the projectile's initial velocity and the (vertical) acceleration of gravity. That tells you when the projectile hits the plane. If you want to know where the projectile hits the plane you can use the (known, constant) horizontal velocity of the projectile and the elapsed time to figure out how far it traveled horizontally.

But the reason this works is not because the vertical components in the first part of the analysis are vertical. It works because they are perpendicular to the plane, and everything else is parallel to the plane.

In the "inclined" frame of reference, you have both a perpendicular and a parallel component of gravity. The parallel component will cause the calculation of distance traveled parallel to the plane to be more "interesting" (by which I mean, more complicated) than it was for the non-inclined plane. But the point is, if you separately consider two components of position of the projectile, one perpendicular to the plane and one parallel, the parallel components of velocity and acceleration affect only the parallel component of position (just as in the level-plane case, where the parallel acceleration happened to be zero) and the perpendicular components of velocity and acceleration affect only the perpendicular component of position.

So do the "when does the projectile hit the ground" calculation for the inclined plane exactly as you would for the level plane, but make sure that you are using components of acceleration and velocity measured perpendicular to the plane, not the vertical components.