Projectiles on an Inclined Plane

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. You stand at the bottom of a hillside and throw a ball up the slope, as far up as possible. On flat ground, you know the best angle: forty five degrees. But on a slope? Is forty five degrees still the best? Not so obvious, right? Today we find out. And in JEE, the inclined plane is where projectile questions get interesting. The whole trick is one smart choice of axes. So, let's begin.

0:33Tilt the axes: along and across the incline

So, look at the figure. The grey wedge is our incline, and alpha, this Greek letter at the bottom, is its angle with the horizontal. The orange arrow, u with a small arrow on top, is u vector, the velocity of projection. And theta is its angle measured above the incline, not above the ground. Remember this, okay? The blue curve is the path, landing back on the incline. Usually we take x horizontal and y vertical. But here the ball starts and lands on the slope, right? So tell me, which directions should our axes take? Yes, exactly! Along the incline, and across it. In the figure, the grey arrows: x along the incline, up the slope, and y across it, at right angles to the slope. Such a natural choice. But gravity, the green arrow g, points straight down, along neither axis. So we resolve it. Along the incline, a sub x, the a with a small x below, equals minus g sin alpha. Minus, because it pulls the ball down the slope, while x points up. Across, a sub y equals minus g cos alpha, into the incline. The corner of the figure shows these too, with the time of flight, which we find next. Now it is just two one dimensional motions. With me?

1:55Time of flight: the motion across the incline

So, the time of flight. Look only across the incline, along y. The ball starts with u sin theta, the part of u at right angles to the incline, and its acceleration is minus g cos alpha. Like a ball thrown straight up, with a smaller g. Now tell me: when the ball lands, what is its y? Zero, yes! y is the distance from the incline, and the ball lands back on the incline. So y equals zero again. Solve it like a ball thrown up, and you get the time of flight. T equals two u sin theta, by g cos alpha. Write it down. Now compare with flat ground. There, T was two u sin theta by g. Here, the only change: g becomes g cos alpha. That's it! Clear?

2:46Range along the incline, up and down

Next, the range: how far along the incline the ball lands. Put this time into the motion along the incline, simplify, and you get this. R up, the small word up means up the plane, equals two u squared sin theta cos of theta plus alpha, by g cos squared alpha. And notice, theta plus alpha is just the angle of the throw from the horizontal, right? Alright, and the most useful one: the maximum range up the plane. R up max, the small max means maximum, equals u squared, by g into one plus sin alpha. Now a quick check. Make the slope flat: put alpha equal to zero. What does this formula become? Sin zero is zero, so it becomes u squared by g. Yes, exactly the flat ground maximum! The formula passes the test. Now throw down the slope instead. Nothing new to derive: just replace alpha by minus alpha. So before I write it, you guess. Will the maximum range down the slope be bigger or smaller than up the slope? Look: R down max equals u squared, by g into one minus sin alpha. One minus sin alpha is smaller than one plus sin alpha, so we divide by a smaller number. Bigger range! Got it?

4:08Hitting the incline at right angles

Last result. Sometimes the ball strikes the incline at right angles. Think: at that moment, what is its velocity along the incline? Zero, correct! Hitting at right angles means the whole velocity is across the incline. The double arrow means so: v sub x equals zero, at landing. Along the incline, the ball starts with u cos theta and slows down at g sin alpha. So u cos theta, minus g sin alpha into T, equals zero. And T is our time of flight, two u sin theta by g cos alpha. Put T in. The u cancels, the g cancels, then multiply through by cos alpha, and we get: cos theta cos alpha equals two sin theta sin alpha. Following so far? Divide both sides by sin theta cos alpha, and that's it: cot theta equals two tan alpha. Cot is one by tan, okay? Remember this one line.

5:09Example 1: time of flight and range up a 30 degree incline

Let's do example one. So, a particle is projected at ten metres per second, that is the m s to the minus one on screen, at thirty degrees above an incline of thirty degrees. The small circle means degrees, okay? So it goes at sixty degrees to the horizontal, up the plane. Find the time of flight and the range along the incline. Take g as ten metres per second squared. Pause the video, and try it first. Done? Okay, let's check. The ball goes at sixty degrees to the horizontal, so theta is sixty, right? No, no, careful! This is the classic trap. Theta is measured above the incline. So theta is thirty degrees, and alpha is also thirty. Sixty is theta plus alpha, okay? Time of flight. On top: two into ten into sin thirty, which is half. The dot means multiply. Below: ten into cos thirty, root three by two. That gives two by root three, about one point one five seconds. The wavy equals sign means approximately. Now the range. On top: two, into u squared, a hundred, into sin thirty, half, into cos of theta plus alpha, cos sixty. Below: ten into cos squared thirty, three by four. Fifty by seven point five: twenty by three metres. Nice and clean!

6:35Example 2: maximum range up a 30 degree incline

Now a harder one, and it answers our puzzle. Find the maximum range up an incline of thirty degrees, for a speed of twenty metres per second, and the direction of projection for it. Take g as ten. Pause and try. Hint: use the maximum range formula. Done? Okay, let's check. R up max equals u squared, by g into one plus sin alpha. u squared is four hundred, and sin thirty is half. So, four hundred by ten into one plus half. Ten into one and a half is fifteen. Four hundred by fifteen: eighty by three, about twenty six point seven metres. Now the direction, our puzzle. On flat ground, the best was forty five degrees. So here also, aim at forty five degrees to the horizontal, right? No, no, careful! On an incline, the best direction is forty five degrees minus alpha by two, measured above the incline. Alpha is thirty, so forty five minus half of thirty: thirty degrees above the incline. And from the horizontal? Theta plus alpha: thirty plus thirty, sixty degrees. Not forty five! And that answers our hillside puzzle: on this slope, the best throw is steeper than forty five. Makes sense?

7:58Quick revision: T, maximum range, hitting at right angles

So, quick revision. Axes along and across the incline: a sub x is minus g sin alpha, a sub y is minus g cos alpha, and T is two u sin theta by g cos alpha. Maximum range: u squared, by g into one plus sin alpha, up the plane, and one minus sin alpha, down. And hitting at right angles means v sub x is zero on landing, so cot theta equals two tan alpha. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!