- Horizontally launched projectile
- What is 2D projectile motion?
- Visualizing vectors in 2 dimensions
- Projectile at an angle
- Launching and landing on different elevations
- Total displacement for projectile
- Total final velocity for projectile
- Correction to total final velocity for projectile
- Projectile on an incline
- 2D projectile motion: Identifying graphs for projectiles
- 2D projectile motion: Vectors and comparing multiple trajectories
- What are velocity components?
- Unit vectors and engineering notation
- Unit vector notation
- Unit vector notation (part 2)
- Projectile motion with ordered set notation
How to solve for the horizontal displacement when the projectile starts with a horizontal initial velocity. We also explain common mistakes people make when doing horizontally launched projectile problems. Created by David SantoPietro.
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- Is acceleration due to gravity 10 m/s^2 or 9.8 m/s^2? My teacher says it is 10 but Dave says it is 9.8.(0 votes)
- Well, if nothing is mentioned then use 10m/s^2. It makes your calculation easier. But look out for values like 4.9 or 49 or 9.8 or 98 in your question. If you get values like that in your question then go ahead and use 9.8m/s^2(6 votes)
- Why does the time remain same even if the body covers greater distance when horizontally projected? I mean when the body is just dropped without any horizontal component, it will fall straight. But when we give a horizontal velocity to the body, it should cover a parabolic path(greater than the path covered during free fall). So the body should take a longer time to fall.
: )(17 votes)
- The acceleration due to gravity is the same whether the object is falling straight or moving horizontally. Since acceleration is the same, then the time each object hits the ground will be the same, assuming they both start from the same height and fall the same distance.(29 votes)
- David mentioned that the time it takes for vertical displacement to occur would the same as the time it takes for the horizontal displacement to happen. This much makes sense, especially if air resistance is negligible.
However, what happens in the case of a cliff jumper with a wing suit? Are the times still the same for the vertical and horizontal?(14 votes)
- If in a horizontally launched projectile problem you're given the height of the 'cliff' and the horizontal distance at which the object falls into the 'water' how do you calculate the initial velocity?(3 votes)
- Maths version of what Teacher Mackenzie said:
Find the time it takes for an object to fall from the given height.
∆y = v_0 t + (1/2)at^2; v_0 = 0; ∆y = -h; and a = g the initial vertical velocity is zero, because we specified that the projectile is launched horizontally.
-h = (1/2)gt^2
-2h/g = t^2
√(-2h/g) = t The negative sign under the radical is fine because gravitational acceleration is also in the negative direction.
Then we take this t and plug it into the x equations
∆x = v_0t + 1/2at^2; horizontal acceleration is zero. We could also use an equation with final velocity instead of acceleration, using the understanding that final velocity will equal initial velocity.
∆x = v_0*t; solve for initial velocity
∆x/t = v_0(3 votes)
- If something is thrown horizontally off a cliff, what is it’s vertical acceleration? How do you know? What is its horizontal acceleration? How do you know?(2 votes)
- Its vertical acceleration is -9.8 m/(s^2) (the acceleration due to gravity) and a projectile (if you're neglecting air resistance) never has acceleration in the horizontal direction. So in the horizontal direction the acceleration would be 0. Hope this helps!(7 votes)
- How would you then find the velocity when it hits the ground and the length of the hypotenuse line?(2 votes)
- If you were asked to find final velocity, you would need both the vertical and horizontal components of final velocity. Horizontal is easy, there is no horizontal acceleration, so the final velocity is the same as initial velocity (5 m/s). To find the vertical final velocity, you would use a kinematic equation. You have vertical displacement (30 m), acceleration (9.8 m/s^2), and initial velocity (0 m/s). You could then use the time-independent formula:
Vf^2 - Vi^2 = 2 * a * d
Vf^2 - (0)^2 = 2 * (9.8) * (30)
Vf = sqrt(2 * 9.8 * 30)
Vf = 24.2...
Now that you have the final velocity components, you can set up a right triangle to solve for the combined final velocity. The components will be the legs, and the total final velocity will be the hypotenuse. By the pythagorean theorem:
Vfx^2 + Vfy^2 = Vf^2
(5)^2 + (24)^2 = Vf^2
Vf = sqrt(5^2 + 24.2^2)
Vf = 24.7
That's the magnitude of the final velocity. To find the angle, you would need to do some trig and realize that the angle from the horizontal is opposite to Vfy and adjacent to Vfx. We can write this as:
tan(theta) = Vfy / Vfx
theta = atan ((24.2) / (5))
theta = 78.3... degrees
And there you have both the magnitude and angle of the final velocity. Hope this helps! (Yes, I am the slightest bit too lazy to actually write the symbol for theta)(4 votes)
- Would air resistance shorten the horizontal distance you are jumping, or lengthen it?(2 votes)
- But what if you are given initial velocity, say shot from a canon, and asked to find the x and the y components and the angle? You are given the displacement in x and a time so can you still assume acceleration in the x is 0?(2 votes)
- 9:18whre did he get that formula,? in the delta y formula is asking to elevate to 2 now doing the root he is decreasing, i dont catch it(1 vote)
- It's not a formula. It's simple algebra.
-30 = 0 + 1/2 * (-9.8) * t^2
-30 = 1/2 * (-9.8) * t^2
Multiply both sides of the equation by 2,
-30 * 2 = (two divided by 2 results into 1) * (-9.8) * t^2
-60 = (-9.8) * t^2
Divide by -9.8 on both sides
-60/-9.8 = t^2.
And then take square root for t and solve.(5 votes)
- gravity should not influence the x-velocity, but that's under the assumption that gravity in uniform and only pulls downward. This is only true if the earth was flat, but of course it is not.
Thus, shouldn't gravity have an impact on the x-velocity in real life, no matter how negligible?(3 votes)
- [Instructor] Let's talk about how to handle a horizontally launched projectile problem. These, technically speaking, if you already know how to do projectile problems, there is nothing new, except that there's one aspect of these problems that people get stumped by all of the time. So I'm gonna show you what that is in a minute so that you don't fall into the same trap. What we mean by a horizontally launched projectile is any object that gets launched in a completely horizontal velocity to start with. So if something is launched off of a cliff, let's say, in this straight horizontal direction with no vertical component to start with, then it's a horizontally launched projectile. What could that be? I mean a boring example, it's just a ball rolling off of a table. If you just roll the ball off of the table, then the velocity the ball has to start off with, if the table's flat and horizontal, the velocity of the ball initially would just be horizontal. So if the initial velocity of the object for a projectile is completely horizontal, then that object is a horizontally launched projectile. A more exciting example. People do crazy stuff. Let's say this person is gonna cliff dive or base jump, and they're gonna be like "whoa, let's do this." We're gonna do this, they're pumped up. They're gonna run but they don't jump off the cliff, they just run straight off of the cliff 'cause they're kind of nervous. Let's say they run off of this cliff with five meters per second of initial velocity, straight off the cliff. And let's say they're completely crazy, let's say this cliff is 30 meters tall. So that's like over 90 feet. That is kind of crazy. So 30 meters tall, they launch, they fly through the air, there's water down here, so they initially went this way, and they start to fall down, and they do something like pschhh, and then they splash in the water, hopefully they don't hit any boats or fish down here. That fish already looks like he got hit. He or she. Alright, fish over here, person splashed into the water. We want to know, here's the question you might get asked: how far did this person go horizontally before striking the water? This is a classic problem, gets asked all the time. And if you were a cliff diver, I mean don't try this at home, but if you were a professional cliff diver you might want to know for this cliff high and this speed how fast do I have to run in order to avoid maybe the rocky shore right here that you might want to avoid. Maybe there's this nasty craggy cliff bottom here that you can't fall on. So how fast would I have to run in order to make it past that? Alright, so conceptually what's happening here, the same thing that happens for any projectile problem, the horizontal direction is happening independently of the vertical direction. And what I mean by that is that the horizontal velocity evolves independent to the vertical velocity. Let me get the velocity this color. So say the vertical velocity, or the vertical direction is pink, horizontal direction is green. This vertical velocity is gonna be changing but this horizontal velocity is just gonna remain the same. These do not influence each other. In other words, this horizontal velocity started at five, the person's always gonna have five meters per second of horizontal velocity. So this horizontal velocity is always gonna be five meters per second. The whole trip, assuming this person really is a freely flying projectile, assuming that there is no jet pack to propel them forward and no air resistance. This person's always gonna have five meters per second of horizontal velocity up onto the point right when they splash in the water, and then at that point there's forces from the water that influence this acceleration in various ways that we're not gonna consider. How about vertically? Vertically this person starts with no initial velocity. So this person just ran horizontally straight off the cliff and then they start to gain velocity. So they're gonna gain vertical velocity downward and maybe more vertical velocity because gravity keeps pulling, and then even more, this might go off the screen but it's gonna be really big. So a lot of vertical velocity, this should keep getting bigger and bigger and bigger because gravity's influencing this vertical direction but not the horizontal direction. So how do we solve this with math? Let's write down what we know. What we know is that horizontally this person started off with an initial velocity. V initial in the x, I could have written i for initial, but I wrote zero for v naught in the x, it still means initial velocity is five meters per second. And we don't know anything else in the x direction. You might think 30 meters is the displacement in the x direction, but that's a vertical distance. This is not telling us anything about this horizontal distance. This horizontal distance or displacement is what we want to know. This horizontal displacement in the x direction, that's what we want to solve for, so we're gonna declare our ignorance, write that here. We don't know how to find it but we want to know that we do want to find so I'm gonna write it there. How about in the y direction, what do we know? We know that the, alright, now we're gonna use this 30. You might want to say that delta y is positive 30 but you would be wrong, and the reason is, this person fell downward 30 meters. Think about it. They started at the top of the cliff, ended at the bottom of the cliff. It means this person is going to end up below where they started, 30 meters below where they started. So this has to be negative 30 meters for the displacement, assuming you're treating downward as negative which is typically the convention shows that downward is negative and leftward is negative. So if you choose downward as negative, this has to be a negative displacement. What else do we know vertically? Well, for a freely flying object we know that the acceleration vertically is always gonna be negative 9.8 meters per second squared, assuming downward is negative. Now, here's the point where people get stumped, and here's the part where people make a mistake. They want to say that the initial velocity in the y direction is five meters per second. I mean people are just dying to stick these five meters per second into here because that's the velocity that you were given. But this was a horizontal velocity. That's why this is called horizontally launched projectile motion, not vertically launched projectile motion. So think about it. The initial velocity in the vertical direction here was zero, there was no initial vertical velocity. This person was not launched vertically up or vertically down, this person was just launched straight horizontally, and so the initial velocity in the vertical direction is just zero. People don't like that. They're like "hold on a minute." They're like, this person is gonna start gaining, alright, this person is gonna start gaining velocity right when they leave the cliff, this starts getting bigger and bigger and bigger in the downward direction. But that's after you leave the cliff. We're talking about right as you leave the cliff. That moment you left the cliff there was only horizontal velocity, which means you started with no initial vertical velocity. So this is the part people get confused by because this is not given to you explicitly in the problem. The problem won't say, "Find the distance for a cliff diver "assuming the initial velocity in the y direction was zero." Now, they're just gonna say, "A cliff diver ran horizontally off of a cliff. "Find this stuff." And you're just gonna have to know that okay, if I run off of a cliff horizontally or something gets shot horizontally, that means there is no vertical velocity to start with, I'm gonna have to plug this initial velocity in the y direction as zero. So that's the trick. Don't fall for it now you know how to deal with it. So we want to solve for displacement in the x direction, but how many variables we know in the y direction? I mean we know all of this. This is good. But we can't use this to solve directly for the displacement in the x direction. We need to use this to solve for the time because the time is gonna be the same for the x direction and the y direction. So I find the time I can plug back in over to there, because think about it, the time it takes for this trip is gonna be the time it takes for this trip. It doesn't matter whether I call it the x direction or y direction, time is the same for both directions. In other words, the time it takes for this displacement of negative 30 is gonna be the time it takes for this displacement of whatever this is that we're gonna find. So let's solve for the time. Now, how will we do that? Think about it. We know the displacement, we know the acceleration, we know the initial velocity, and we know the time. But we don't know the final velocity and we're not asked to find the final velocity, we don't want to know it. So let's use a formula that doesn't involve the final velocity and that would look like this. So if we use delta y equals v initial in the y direction times time plus one half acceleration in the y direction times time squared. Alright, now we can plug in values. My displacement in the y direction is negative 30. My initial velocity in the y direction is zero. This is where it would happen, this is where the mistake would happen, people just really want to plug that five in over here. But don't do it, it's a trap. So, zero times t is just zero so that whole term is zero. Plus one half, the acceleration is negative 9.8 meters per second squared. And then times t squared, alright, now I can solve for t. I'm gonna solve for t, and then I'd have to take the square root of both sides because it's t squared, and what would I get? I'd have to multiply both sides by two. So I get negative 30 meters times two, and then I have to divide both sides by negative 9.8 meters per second squared, equals, notice if you would have forgotten this negative up here for negative 30, you come down here, this would be a positive up top. You'd have a negative on the bottom. You'd have to plug this in, you'd have to try to take the square root of a negative number. Your calculator would have been all like, "I don't know what that means," and you're gonna be like, "Er, am I stuck?" So you'd start coming back here probably and be like, "Let's just make stuff positive and see if that works." It would work because look at these negatives canceled but it's best to just know what you're talking about in the first place. So be careful: plug in your negatives and things will work out alright. So if you solve this you get that the time it took is 2.47 seconds. It's actually a long time. It might seem like you're falling for a long time sometimes when you're like jumping off of a table, jumping off of a trampoline, but it's usually like a fraction of a second. This is actually a long time, two and a half seconds of free fall's a long time. So we could take this, that's how long it took to displace by 30 meters vertically, but that's gonna be how long it took to displace this horizontal direction. We can use the same formula. We can say that well, if delta x equals v initial in the x direction, I'm just using the same formula but in the x direction, plus one half ax t squared. So the same formula as this just in the x direction. Delta x is just dx, we already gave that a name, so let's just call this dx. So I'm gonna scooch this equation over here. Dx is delta x, that equals the initial velocity in the x direction, that's five. Alright, this is really five. In the x direction the initial velocity really was five meters per second. How about the initial time? Oh sorry, the time, there is no initial time. The time here was 2.47 seconds. This was the time interval. The time between when the person jumped, or ran off the cliff, and when the person splashed in the water was 2.4, let me erase this, 2.47 seconds. So 2.47 seconds, and this comes over here. How about this ax? This ax is zero. Remember there's nothing compelling this person to start accelerating in x direction. If they've got no jet pack, there is no air resistance, there is no reason this person is gonna accelerate horizontally, they maintain the same velocity the whole way. So what do we get? If we solve this for dx, we'd get that dx is about 12.4, I believe. Let's see, I calculated this. 12.4-ish meters. Okay, so if these rocks down here extend more than 12 meters, you definitely don't want to do this. I mean if it's even close you probably wouldn't want do this. In fact, just for safety don't try this at home, leave this to professional cliff divers. I'm just saying if you were one and you wanted to calculate how far you'd make it, this is how you would do it. So, long story short, the way you do this problem and the mistakes you would want to avoid are: make sure you're plugging your negative displacement because you fell downward, but the big one is make sure you know that the initial vertical velocity is zero because there is only horizontal velocity to start with. That's not gonna be given explicitly, you're just gonna have to provide that on your own and your own knowledge of physics.