 A bullet is fired from a gun and a second bullet is just dropped at the same time the first bullet is fired. And this is purely fired horizontally. The question is to figure out which bullet will hit the ground first. Will it be this one or will it be this one? This is a question that we will answer in this video, but before we do that we will jump into analyzing projectile motion, we will try to understand how to solve any projectile motion problem mathematically. And as we do that, we will come across some new words such as range, we will try to understand what range is, we will also drive a formula for maximum height in a projectile motion and also the time it takes for the object to complete its motion. So we will talk about all of this, also how to solve projectile motion problem and then finally come back to this question and answer it. Alright, let's begin. So let's stick to the same object, the bullet over here. So now projectile motion we know, we know that the velocity can be resolved into two components. There is a vertical component, there is a horizontal component and it looks like this. The vertical component of the velocity it keeps on decreasing, becomes zero at the topmost point, then keeps on increasing as it goes down and the horizontal component stays the same, nothing happens to it. That is because there is an acceleration due to gravity that is acting in the downward direction so that is only changing the vertical component of the velocity and there is no acceleration in the x direction, the horizontal direction. So nothing happens to the velocity in the horizontal direction. Now one way to talk about projectile motion in all its detail is, that is it's to really treat x and y separately. The horizontal motion and the vertical motion treat them separately. So if we say that the bullet was fired with a velocity of u at an angle of theta, we can resolve this into components. The horizontal component, this would be u cos theta. The component adjacent to theta is always cos and the component opposite to it, opposite to theta is the sine component. So this is the initial vertical velocity that is u sine theta. So what all do we know in x? We know that there is no acceleration, so ax is zero. The initial, the velocity in the x direction that remains the same throughout. So ux, ux that is really just u cos theta. And that's it, we don't know anything else. For the y, for the y direction, we know the vertical component, that is the initial vertical component and also the acceleration that is involved in the motion. So ay, ay this is really minus g. If we take the upward direction as positive, this g is acting downwards, so that will be negative. And we know uy, that is the initial vertical component of the velocity, this is u sine theta. So now what we will do is, we will write the kinematic equations, just so that we can refer to them. So I'm going to write them over here. We have v, this is equals to u plus at, this is one. One more is v square, this is equals to u square plus 2as. And finally s, this is equals to ut plus half at square. So now if we first start to think about time, if we start to think about time, the time that the bullet takes to complete its path, time for the entire projectile motion. We can then ask ourselves which component really is having some influence or determining time. So if you try to imagine these two motions separately, the vertical motion is identical to a motion. For instance, let's say you threw a stone vertically with this velocity of u sine theta. And the horizontal component is identical to the motion of a particle or anything that is moving horizontally. But when you combine these two, you get a projectile, you get velocity at a certain angle. So the time that the bullet stays in the air, that's really just dependent upon the vertical component. If this component was longer, then the bullet would go to a greater height and take more time to fall to the ground. But if u sine theta was shorter, then the bullet will not even reach to this height and fall to the ground much sooner. But if there was no u sine theta, then there would have been just horizontal motion. There wouldn't have been any projectile even to work with. So the time is really just dependent on the vertical motion. It's not at all dependent on horizontal. Because the time that the bullet stays in the air is really just determined by u sine theta. u cos theta plays no role in that. So if we try to figure out the time of the journey, we only look at the y direction. And here we can see that when the bullet falls back to the ground, there is no vertical displacement. It's there at the same level that it began with. There is no vertical displacement. So if we try to think about which equation can we use for time, let's try to understand what all do we really know. We know that the vertical displacement to figure out time, to figure out time, we know that the vertical displacement, this is, this is zero. There is no vertical displacement because it's starting from this point and it's ending at this point. There is no vertical displacement. There was a positive then equal negative. And what else do we know? We know the initial vertical velocity, u sine theta. This is u sine theta. And we also know the acceleration that is minus g. So if you look at this third equation, we know s and we know this part, ut. And we also know a. Only one variable remains at its t. So we can use this. So when we do s is just zero, vertical displacement is zero. So zero, this is equal to ut. This is u sine theta into t minus half gt square. That is because ay is minus g. When we work this out, we can take t common in this, in this expression. And that gives us u sine theta minus half gt. And that is equal to zero. So two solutions will come out from this, this equation. One would be that t is equal to zero. And the other would be t being equal to two u sine theta divided by g. Now t equals to zero. It really tells us, tells us the position of the bullet when it was just fired from this point. That is t equals to zero. t equals to two u sine theta by g really tells us the time when the bullet is, bullet is at this position, at this position right here. Now we can try and describe this motion more. We can, we can, we can talk about, we can talk about the horizontal displacement. And we call this range. This is called, this is called range. So this is the second thing that we are talking about. Range is the horizontal displacement. And when we talk about horizontal displacement, I'm sure you'll agree that we only need to look at the x part of the motion. Because horizontal displacement is only dependent on how long or short the blue vector is. It does not really depend on u sine theta. You can have u sine theta extremely large or extremely small. It still won't play any role on how much the bullet moved in the x direction, in the horizontal direction. And usually we ignore air resistance so there is no acceleration. So to figure out displacement, we know that speed is equals to distance upon times and velocity is displacement upon time. So displacement range that is equal to velocity, which is u cos theta into, into time. And we already calculated time. We already calculated time over here. So we can really place this over here. And when we do that, this becomes equal to two u square sine theta into cos theta divided by hg. This is your horizontal displacement. And we only looked at x to be able to figure this out. But we also needed to know what time was, right? And for time, we looked at, we looked at the vertical, we looked at the vertical motion. So now one more thing remains to be able to fully describe the projectile motion. That is the maximum height. That is this maximum height. Now at the maximum height, at this point, at this point right here, there is no vertical velocity. There is only, only a horizontal velocity. And the component that influences maximum height is only the vertical component. It depends how long or short this vector is. That will only tell us how high or how low the projectile goes. Horizontal component only determines the horizontal displacement, not the vertical displacement. Vertical displacement is the, is u sine theta's effect. So, so at the highest point, there is no vertical velocity. There is no vertical component of velocity. So now if we try and think about which equation can we use to figure out, let's call this H to figure out H. Let's, let's think about it. We know the, we know the initial vertical velocity and we know acceleration. We know acceleration. We know this. And we also know the final vertical velocity. So this really, this really, this is i, u, y, f. And by final, I mean the velocity at the highest point. This is zero. So now this equation really makes sense because we know v, u and we know a, we need to figure out s. So when we do that, when we do that, this is v square. The final vertical component of velocity at this point, at the highest point, this is zero. This is zero. u square would be u square sine square theta and a is minus g. So when we take u square to the left hand side, that becomes u square sine square theta. Let me just add the minus sign also. And this is equal to minus 2g H minus gets cancelled off and H becomes equal to u square sine square theta divided by 2g. And this is the third thing. This is the maximum height. So to be able to describe a projectile motion, all we need to know is the velocity with which the object was fired and the angle at which it was fired. We can then tell the time, maximum height and the horizontal displacement or the range. And to be able to tell maximum height, we only looked at the vertical part of the motion. To be able to tell the horizontal displacement, the range, we only looked at the horizontal part of the motion. And to figure out time, we again looked at the vertical, the vertical part of the motion. Now let's go back to the question that we asked in the beginning of this video. All right. So between these two bullets, the question is which bullet will hit the ground first. And this brown bullet is being shot horizontally, it's totally horizontal. So when it is shot, there is no vertical component to the velocity, there is just horizontal component. And the vertical component builds because of gravity. And the red bullet is just dropped at this point, there is only a vertical component. There is no horizontal component to this velocity. Now why don't you pause the video and maybe try to think which bullet will hit the ground first. Okay, let's try and change the question a little. So what will you say, what will you say if the gun was angled at this point and the bullet was fired and at that instant, a second bullet was dropped vertically. In that case, which bullet do you think will hit the ground first, the one that was fired at an angle or the one that was just dropped at that very instant. I think you might have guessed it right, the one that was dropped at that very instant, right? Why don't you, why don't you take a guess at this point. Now when a bullet is dropped at this point and when a bullet is fired, which one will hit the ground first? Here you might say the other bullet, the brown bullet, this bullet will hit the ground first and the brown bullet. Okay, what about, what about in this case, it is angled slightly like this. So the bullet goes somewhat like this, the brown bullet goes like this and the other bullet is just dropped at that very instant. Now again you might say the red bullet will, will reach the ground first. And now if I tilt it slightly like this, now you might say the brown bullet will hit the ground first. So there comes a point, there comes a point when both of these bullets, they hit the ground at the same time and that is when the bullet is fired horizontally. The bullet that is fired horizontally and the bullet that is dropped, they reach the ground at the same time. Let's try to confirm that mathematically. So this height, we can call that, we can call that h and let's try to figure out the time that both of them take to reach the ground. So for the red one, for the red one, initial velocity that is zero, then it gains some velocity because of acceleration due to gravity. And the time that it takes, we can use a kinematic equation which is s is equals to half at square and over here, s is h, this h, this is equal to, we could have written ut but initial, initial velocity is zero. It has just been dropped. So initial velocity, this is zero. So h, this is equal to half gt square. And t is really the under root of, this is 2h divided by g. And now if you look at the brown bullet, if you look at the brown bullet, here if you look at only the y part of the journey, initially there is no velocity. At this point there is some, at this point there is more, at this point there is even more, at this point even more. So the vertical motion of this bullet is just like the vertical motion of this bullet. Even for this bullet, velocity is increasing, increases some more, some more, some more and then more. Now if you try to figure out the time that the brown bullet takes to reach the ground, we only really need to just look at the y component. Initial velocity, initial vertical velocity is zero. So just like this, it's zero. And it falls to the same height. So h is again equal to half at square plus ut, ut, this is zero because there is no vertical component of velocity when it is just fired horizontally. There is only a horizontal component. So now again t, this is equal to under root of 2h divided by g, in place of a, we can write g. So these two bullets, they reach the ground at the same time, mainly because they both start with zero vertical velocity, their vertical velocity increases because of the same acceleration that is g and they fall through the same height that is h. So they reach the ground at the same time. It doesn't matter how fast you fire it really because the vertical velocity will still be zero when it is shot at a greater speed horizontally. It will still be zero. And the time that it takes to reach the ground, the time that it takes to cover this height edge is only dependent on the vertical velocity. We're not looking anything horizontal. We're only looking at how much time it takes for the bullet to complete this distance edge that is only dependent on the vertical component. Vertical distance dependent on vertical component. And that starts from zero increases because of g, just like the red bullet. So which bullet will hit the ground first? Both reach the ground at the same time.