Velocity at an Instant

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Two friends throw two balls from the same spot, with the same speed: one steeply upwards, the other at a lower angle. Both balls fly past the same window. Which ball is moving faster at the window? The steep one? Hmm, keep your answer ready. Today we find the velocity of a projectile at any instant: its speed and its direction. This idea is very useful for JEE problems. So, let's begin.

0:32Split the velocity into two parts

So, look at the picture, drawn on x and y axes. The ball is launched with speed u at an angle theta; the arrow on u makes it u vector. H, R and R by two are from the last lesson, and the speed u at landing we will see soon. At every instant, we split the velocity into two parts: v x along the ground, in green, and v y straight up, in orange. The small x and y below the v give the direction. Now think. Gravity pulls only downwards. So which part does it change? Yes, only v y! Along the ground nothing changes, so v x equals u cos theta, constant, for the whole flight; even the green arrow at the top says u cos theta. Okay? And v y starts as u sin theta, and gravity takes away g every second. So v y equals u sin theta minus g t, where t is the time since launch. Exactly like a ball thrown straight up, right?

1:32Speed and direction at any instant

Now the full velocity. The two parts are perpendicular, so the speed v is the square root of v x squared plus v y squared. The tick-shaped sign is the square root, okay? And the direction? Let phi, this Greek letter, be the angle the velocity makes with the horizontal. Then tan phi equals v y by v x. Look, as the ball rises, v y shrinks, so the velocity turns flatter and flatter. Clear? Alright, now a classic JEE trap. At the very top, the ball stops going up. So its velocity there is zero, right? No, no, careful! At the top, only v y is zero. v x is still u cos theta, so the ball moves horizontally, with speed u cos theta, like the green arrow in our picture. Got it?

2:23Speed at a height h

Now, back to our puzzle. Same launch speed u, different angles, the same height h. Does the speed at that height depend on theta? Let's find out. v squared equals v x squared plus v y squared. And from straight-line motion, vertically, v y squared equals u squared sin squared theta, minus two g h. With me? Put in v x equals u cos theta. So v squared equals u squared cos squared theta, plus u squared sin squared theta, minus two g h. Take u squared common. Cos squared theta plus sin squared theta is one, right? So v squared equals u squared minus two g h. See? Theta has disappeared! So the speed at height h is the square root of u squared minus two g h, whatever the angle. Puzzle solved! Did you say the steep one? Actually, no: at the window, both balls have exactly the same speed. Exactly the same! Only their directions differ.

3:26Over the whole flight

Alright, now the whole flight. Put h equals zero in our formula: the ball lands with speed u, the same as at launch. Same speed at the start and the end, so the change in velocity is zero, right? No, no! Velocity is a vector; direction matters. v x does not change, but v y goes from u sin theta, upwards, to minus u sin theta, downwards. The small arrow means becomes. So the change in velocity is minus two u sin theta, j cap. The triangle, delta, means change; the arrow makes it v vector; and j cap is the unit vector pointing up. So the minus sign means straight down. Clear? Now the average velocity: displacement divided by time. So what is the displacement from launch to landing? Yes, just the range R, along the ground, in the time of flight T, okay? So the average velocity is R by T, along i cap. And R by T is u squared sin two theta by g, over two u sin theta by g. Write sin two theta as two sin theta cos theta, and everything cancels down to u cos theta. So the average velocity is u cos theta, i cap; i cap is the unit vector along the ground. Look, that is just v x! So, topper's trick: over the whole flight, the average velocity is simply u cos theta, i cap, no need to find R and T at all. Got it?

4:58Example 1: velocity after 1 s for 20 m/s at 60 degrees

Right, let's start with an easy one. Here, a stone is thrown at twenty metres per second, at sixty degrees; the m s to the power minus one means metres per second, and the small circle means degrees. With g as ten metres per second squared, find its velocity after one second. Pause the video, and try it first. Done? Okay, let's check. Gravity takes away ten every second, so after one second the velocity is twenty minus ten into one, ten metres per second, right? The cross on screen means multiply. No, no! That is the trap. Gravity changes only the vertical part. So split first: v x equals twenty cos sixty, twenty into half, ten metres per second. Now v y equals twenty sin sixty, minus ten into one. That is ten root three minus ten: about seven point three metres per second, still going up. The wavy sign means approximately. Clear? Now the speed. v x squared is a hundred, and v y squared is about fifty three point six. So v is root of a hundred plus fifty three point six: about twelve point four metres per second. And the direction: tan phi equals seven point three by ten, zero point seven three. The double arrow means so: phi is tan inverse of zero point seven three, about thirty six degrees above the horizontal; the minus one up there means inverse. Already flatter than sixty, right? That's it, done!

6:34Example 2: velocity perpendicular to the initial velocity

Now a harder one, JEE level. When is the velocity of a projectile perpendicular to its initial velocity? And for which angles does this happen during the flight? Pause and try. Hint: use the dot product. Done? Okay, let's check. Two vectors are perpendicular when what happens? Yes, their dot product is zero! v vector is the velocity at time t, u vector the launch velocity, and the dot between them is the dot product. Multiply the x parts, multiply the y parts, and add. The x parts give u squared cos squared theta. The y parts are u sin theta, and u sin theta minus g t. Okay? Open the bracket: u squared cos squared theta plus u squared sin squared theta is just u squared. So u squared minus u g t sin theta equals zero, and so t equals u over g sin theta. So at this time, the velocity is perpendicular to the launch velocity. Easy! This happens for every angle, right? Wait, careful! This is the trap. The ball is in the air only for the time of flight, two u sin theta by g. So our time must be less than or equal to that: the sign with a line under it, okay? Cancel u and g, and cross multiply: sin squared theta is greater than or equal to half. So sin theta is at least one by root two, and theta at least forty five degrees. But not ninety, okay? At ninety the ball goes straight up, and at this time it is at the top, where the velocity is just zero; zero is not perpendicular to anything. So: t equals u over g sin theta, only for theta from forty five degrees up to, but not including, ninety; that is the sign without the line under it.

8:29Quick revision: vy, speed at height h, change in velocity

Got it? So, quick revision. Only v y changes: v y equals u sin theta minus g t, and v x stays u cos theta. Speed at height h: the square root of u squared minus two g h, whatever the angle. Whole flight: change in velocity, minus two u sin theta, j cap; average velocity, u cos theta, i cap. Now go and try the practice problems on this page. All the best!