Horizontal Projection and Projection from a Height

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. You stand on a rooftop with two stones. One you throw gently, sideways. The other you throw hard, also sideways, from the same height, at the same moment. Which one hits the ground first? The hard one, because it is faster? Or the gentle one, because it travels a shorter path? Today we settle it, with one small formula. And you know, cliffs, tables, planes dropping packets: in JEE, all these questions are built on this one idea. So, let's begin.

0:35Thrown horizontally: two motions, one clock

So, here is our stone, thrown horizontally with speed u, from a height h. In this chapter we always break the motion into two parts, right? Horizontal first. u sub x, the small x below means the horizontal component, is just u. And there is no acceleration sideways, so the sideways speed never changes. The double arrow on screen means "so": x equals u t. Clear? Now the vertical part. At the moment the stone leaves your hand, how much vertical velocity does it have? Zero, exactly! You threw it flat, sideways. So u sub y is zero, and vertically the stone simply falls from rest, through the height h, with the acceleration g. h equals half g t squared, okay? Now solve for t. t equals root of two h by g, the square root of two h over g. This is the time of fall. Write it down; this one line does half the work in every cliff question. Alright, now look at this formula carefully, and remember our two stones. Where is u in it? Nowhere, that's right! Not even once. The time of fall depends only on the height. So the gentle stone and the hard stone hit the ground at the same moment! Puzzle solved. Got it?

1:58How far, how fast, and in which direction

Next: how far out does it land? Horizontal distance is u into the time of fall. x equals u root of two h by g. And here, u is present. So the harder throw lands farther away, but not sooner. Makes sense? Now the velocity on landing. Horizontally, nothing changed: v sub x is still u. Vertically, it started from zero and gained g t. Put in the time of fall, and v sub y becomes root of two g h. So the landing speed is u plus root two g h, right? Just add the two? No, no, careful! This is the classic trap. The two components are at right angles, one horizontal and one vertical. So we combine them like the sides of a right triangle, by Pythagoras. v equals root of u squared plus two g h. Okay? And the direction. Phi, this Greek letter, is the angle the velocity makes with the horizontal. tan phi equals vertical over horizontal: root two g h, by u. And it points below the horizontal, because the stone is coming down. With me?

3:07Dropped from a moving vehicle

Now a favourite JEE situation. Picture a plane flying level, and it drops a packet. Nobody threw the packet; it was just let go. So, seen from the ground, it falls straight down, right? Ah, no! Think again. Just before release, the packet was moving along with the plane, right? So anything dropped from a moving vehicle starts with the vehicle's velocity, at the moment of release. So for a plane flying level, the packet is simply a horizontal projection, with u equal to the plane's speed. All the formulas we just wrote work directly. That's it, nothing new to learn!

3:46Thrown at an angle from a height

Alright, now the harder version. From the same height h, the ball is thrown at an angle, above the horizontal. Put the origin at the launch point, and take upwards as positive. Now tell me: when the ball lands, what is its y? Minus h, yes! The ground is h below the launch point, and upwards is positive, so on the ground, y equals minus h. This minus sign is exactly where most students go wrong. Now the vertical equation. Theta, this Greek letter, is the angle of projection, okay? The vertical starting velocity is u sin theta, and gravity pulls down. So y equals u sin theta t, minus half g t squared, and we set it equal to minus h. This is a quadratic in t; solve it for t. Then the horizontal distance is easy: the horizontal velocity, u cos theta, into t. Got that? Two lines, and any such cliff question is done.

4:48Example 1: thrown horizontally at 10 m/s from a 20 m cliff

Right, let's do example one. So, a stone is thrown horizontally at ten metres per second, that is the m s to the minus one on screen, from a twenty metre cliff. Take g as ten metres per second squared. Find the time of fall, the horizontal distance, and the velocity on landing. Pause the video, and try it first. Done? Okay, let's check. Time of fall first: root of two into twenty, by ten. That is root four, two seconds. And notice, we did not need the speed for this. Only the height, just like our two stones! Horizontal distance: u into t, ten into two. Twenty metres. On landing, v sub x is still ten metres per second. And v sub y is g t, ten into two, twenty metres per second. So the landing speed is ten plus twenty, thirty metres per second, right? No, no! Same trap as before. The components are at right angles, so Pythagoras: root of ten squared plus twenty squared. That is root five hundred, which is ten root five metres per second. And the direction: tan phi equals twenty by ten, which is two. So phi is tan inverse two, below the horizontal. The small minus one up there means inverse: the angle whose tan is two, okay? What a clean answer!

6:17Example 2: thrown at 30 degrees from a 20 m cliff

Now a harder one, JEE level. This time, a ball is thrown at twenty metres per second, at thirty degrees, that small circle on screen means degrees, above the horizontal, from the top of a twenty metre cliff. When does it land, and how far out? Take g as ten. Pause and try. Hint: be careful about where the ground is. Done? Okay, let's check. First, the components. u sub x is twenty cos thirty degrees, ten root three. u sub y is twenty sin thirty degrees, which is ten. Now the vertical equation. u sub y t, minus half g t squared: that is ten t minus five t squared. The ball falls twenty metres, so this equals plus twenty, right? No, careful! Remember the minus sign. The ground is twenty metres below the launch point, so y equals minus twenty. Ten t minus five t squared equals minus twenty. Divide everything by minus five and rearrange, and we get t squared minus two t minus four equals zero. Solve the quadratic: t equals one plus or minus root five. But time cannot be negative, right? So t must be greater than zero, and we keep the plus sign. t equals one plus root five, about three point two four seconds. The wavy equals sign means approximately. Now horizontally: x equals ten root three into t. Ten root three into three point two four: about fifty six metres. See? Same two steps: y equals minus h for the time, then the horizontal motion for the distance. Nice and systematic.

8:04Quick revision: time of fall, landing speed, y = -h

So, quick revision. Thrown horizontally: t equals root of two h by g, and it depends only on the height. Landing speed: combine u and root two g h by Pythagoras, never just add them. And thrown at an angle from a height: set y equals minus h, and solve for t. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!