Projectile Motion from the Ground

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Two fielders throw a cricket ball at the same speed, one at thirty degrees above the horizontal, the other at sixty. Which ball lands farther? The sixty degree one, because it goes higher? Or the thirty? Today we settle it. And remember, projectiles are a favourite topic in JEE papers. So, let's begin.

0:28A projectile: two motions at once

So, first the name. Any body thrown near the earth's surface and then left to gravity, air resistance ignored, is a projectile. In the figure, x runs along the ground and y goes straight up. The orange arrow, u vector, the u with a small arrow on top, is the launch velocity: u is the speed, and theta, its angle with the horizontal. The blue curve is the path. Clear? Gravity pulls straight down. So tell me: what does gravity do to the horizontal motion? Nothing at all! Yes, exactly. u x, the u with a small x below, is the horizontal velocity: u cos theta. It never changes. See the green arrows, v x on the way and u cos theta at the top: same length. Okay? Vertically, u y starts as u sin theta upwards. Gravity gives a constant acceleration g downwards, so this velocity, the orange arrow v y, falls at the rate g. Two simple motions, sharing only the time.

1:34Time of flight, height and range

Alright, now the three numbers almost every projectile question needs. First, the time of flight. The vertical velocity starts at u sin theta upwards. What is it at the moment the ball lands? Yes, u sin theta again, but downwards! See, it lands with the same speed u. The total change is two u sin theta, at the rate g. So T equals two u sin theta by g; the line means divided by. Next, the maximum height. Vertically, it is just a ball thrown straight up at u sin theta. So H equals u squared sin squared theta by two g; sin squared means sin theta, whole squared. The dashed line shows the top comes halfway, at R by two. And the range. The horizontal velocity never changes, so the range is u cos theta into the time of flight; the dot means multiply. Two sin theta cos theta becomes sin two theta, so R equals u squared sin two theta by g. Write all three down, okay?

2:40Maximum range, and two angles with one range

Now, for a fixed speed, only sin two theta can change in the range, right? So for which angle is sin two theta the biggest? Correct, when two theta is ninety degrees, so sin two theta is one. So theta is forty five degrees; the little circle means degrees. And the maximum range is u squared by g. Nice and simple. Alright, back to our puzzle. The sixty degree ball goes much higher, so it must go farther too, right? No, no, careful! For thirty degrees, two theta is sixty; for sixty degrees, it is one hundred and twenty. And sin sixty equals sin one hundred and twenty! So both land at the same spot. Puzzle solved! In general, theta and ninety degrees minus theta give the same range. Got it? Now a topper's trick for such a pair. Call their heights H one and H two; the small one and two label the two throws. H one plus H two equals u squared by two g. The range equals four root of H one H two; the tick-shaped sign is the square root. And T one T two, the times multiplied, equals two R by g. Three lines, and pair questions become one-liners. With me?

3:56The path, and a shortcut for the angle

Next, the path. Remove the time between the x and y motions, and we get: y equals x tan theta, minus g x squared by two u squared cos squared theta. Look, an x squared term, so the path is a parabola. In a neater form: y equals x tan theta, into one minus x by R, in brackets. So y is zero at x equals zero and again at x equals R, both ends of the flight. Keep it ready for Example two. One more shortcut: tan theta equals four H by R, so height and range give the angle in one line. That's it! Now a classic trap. At the very top, the ball stops going up. So its speed there is zero, right? No, no! Only the vertical velocity is zero at the top. The horizontal u cos theta never changes, so that is the speed at the top. And if K is the kinetic energy at launch, at the top it is K cos squared theta. Okay?

5:00Example 1: 20 m/s at 30 degrees, find T, H and R

Now, Example one. Here, a ball is thrown at twenty metres per second, that is m s to the power minus one, at thirty degrees above the horizontal. g is ten metres per second squared. Find the time of flight, the maximum height and the range. Pause the video, and try it first. Done? Okay, let's check. u x equals twenty cos thirty degrees, that is ten root three. u y equals twenty sin thirty degrees, that is ten. Both in metres per second. Time of flight: two u y by g. Two into ten, by ten; the cross means into. So T is two seconds, the s for seconds. Maximum height: u y squared by two g. One hundred by twenty: five metres. Now the range. It flies for two seconds, thrown at twenty metres per second. So the range is twenty into two, forty metres, right? No, no! That is the trap. Only u x, the horizontal part, carries the ball forward. So R equals ten root three into two: twenty root three, about thirty four point six metres; the wavy equals sign means approximately. And by our puzzle, the partner angle at the same speed lands right here too!

6:21Example 2: angle, range, height and speed from the path

Now a harder one, JEE level. The path is y equals two x, minus x squared by ten, in metres, with g ten. Find the angle of projection, the range, the maximum height and the speed. Pause and try. Hint: compare with the path equation. Done? Okay, let's check. Compare with the path equation; the three dots stand for the x squared term. The number with x is tan theta: so tan theta is two, theta is tan inverse two; the small minus one means inverse. For the range, use the neat form: y equals two x, into one minus x by twenty. So y is zero again at x equals twenty: the range is twenty metres. As simple as that. The top comes halfway, at x equals R by two, that is ten. Two into ten, minus one hundred by ten: twenty minus ten. So H is ten metres. Last, the speed. The x squared term has g by two u squared, and here that is one by ten. So u squared is fifty, right? No, no, careful! You forgot the cos squared theta. Leaving out the minus sign, the x squared term has g by two u squared cos squared theta, and that equals one by ten. And cos squared theta is one by one plus tan squared theta, which is one fifth. Clear? Put it in: the left side becomes twenty five by u squared, equal to one by ten. The double arrow means so: u squared is two hundred and fifty. And u is five root ten, about fifteen point eight metres per second. Four answers from one equation. That is the real power of the path!

8:10Quick revision: T, H, R, 45 degrees and speed at the top

So, quick revision. T equals two u sin theta by g, H equals u squared sin squared theta by two g, and R equals u squared sin two theta by g. Maximum range at forty five degrees, and theta and ninety degrees minus theta give the same range. And at the top, the speed is u cos theta, not zero. Now go and try the practice problems on this page. You will find them easy. All the best!