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Impact velocity from given height

Determining how fast something will be traveling upon impact when it is released from a given height. Created by Sal Khan.

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  • blobby green style avatar for user rg2871
    why not just use s=1/2at^2 solve for t than use V=at?
    (13 votes)
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  • duskpin sapling style avatar for user Rohini
    Since the motion of the object and the acceleration (gravity) is in the same direction, aren't you supposed to take + 9.8 m/s^2 ?
    (7 votes)
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  • piceratops ultimate style avatar for user Nathan Shapiro
    How is 9.8 m/sec^2 the affect gravity has on an object? Doesn't it start slower than that and start speeding up until it's faster than that?
    (3 votes)
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    • piceratops sapling style avatar for user Jack Gaines
      The velocity starts at 0 m/s, and then continues to increase as long as the object is falling. So, yes, the object does start slower and will eventually reach speeds faster than 9.8 m/s, but that is velocity and not acceleration. The acceleration remains constant, and is reflecting the increasing velocity of the object as it continues to fall. The object won't stop accelerating until it hits the ground.
      (4 votes)
  • aqualine ultimate style avatar for user Gracie Lynch
    I am confused about the terminology used to describe acceleration and I need some clarification. If you are using a coordinate system where upwards is positive and you throw some thing upward is acceleration positive or negative? If upwards is positive then when you throw something up the acceleration is constantly increasing which to me seems incorrect. I was under the impression that if you throw something upwards the longer it's in the air the lower the acceleration until it starts falling back done and increasing again. Could someone please clarify this for me??
    (3 votes)
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    • male robot hal style avatar for user Charles LaCour
      If you label the up direction as positive and the force of gravity is down then the acceleration due to gravity is negative.

      When you have up as positive and you throw an object up its initial velocity is positive and once it leaves your hand the acceleration of the object is only due to gravity and is constant at -9.8 m/s^2. This vertical acceleration doesn't change.

      As this object travels its initial velocity will change by -9.8 m/s every second. Let's say the object was thrown up at 29.4 m/s. So since the object was thrown up which a positive direction it is initially traveling at + 29.4 m/s.

      After 1 second we know that the velocity changed by - 9.8 m/s so at this point in time the object is traveling at a velocity of (+ 29.4 m/s) + (- 9.8 m/s) = + 19.6 m/s.

      After another second, a total of 2 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be (+ 19.6 m/s) + (- 9.8 m/s) = + 9.8 m/s.

      After another second, a total of 3 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be (+ 9.8 m/s) + (- 9.8 m/s) = 0 m/s.

      After another second, a total of 4 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be ( 0 m/s) + (- 9.8 m/s) = - 9.8 m/s.

      And so on until the object lands.
      (5 votes)
  • spunky sam blue style avatar for user Flattery0427
    If the fall was infinite, could we reach an infinite velocity ?
    (2 votes)
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    • male robot hal style avatar for user Ben Willetts
      With the equations we're using, the answer is yes -- if you had an infinite distance to fall, you would attain an infinite speed. This doesn't happen in real life, though, and the difference is because we are using equations which are slightly simplified. They are accurate enough for everyday use, but they give wrong answers for extreme cases such as "falling an infinite distance". The main differences are that (a) we assume the acceleration (due to gravity) will be constant, but this is only accurate if you're falling distances up to a few km; and (b) we assume that your acceleration doesn't depend on your current speed, but this is only accurate at "everyday" speeds (up to a few hundred km/s) -- in reality it takes more and more force to accelerate the faster you are already going, and it is impossible to accelerate to speeds faster than the speed of light, which is about 300,000 km/s.
      (8 votes)
  • mr pants teal style avatar for user Ali
    For the final question I got -35.64 kilometers per hour and 22.15 miles per hour.
    (4 votes)
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  • leaf green style avatar for user Luqmaan
    How much does air resistance affect the velocity or the acceleration
    (2 votes)
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    • male robot hal style avatar for user Ben Willetts
      Air resistance increases in proportion to the square of the speed of the object, so as the speed increases, the acceleration decreases. There comes a point where acceleration is zero; the object is now at terminal velocity. For a falling human (a sky-diver with closed parachute, for example) this is about 125 mph.
      (3 votes)
  • piceratops ultimate style avatar for user Nikhil
    Sal said we were trying to find the velocity before it hit the ground. Wouldn't the final velocity be when it actually hit the ground?
    (2 votes)
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  • winston default style avatar for user mara.h
    if you throw a rock off that, right at the bottom, right before it hits the ground, it will have a velocity of negative 9.9 meters per second. So negative 9.9 meters per second.
    so if the velocity is any number that will be the ---- per sec?
    (2 votes)
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  • blobby green style avatar for user Noluthando Gasa
    I would like to know how do I go about calculating the angle using the formula R=(V^2sin2theta)/g with an unknown of velocity initial.

    A ball is kicked off the ground reaching a maximum height of 60m and it lands 80m away from the starting point. calculate the initial velocity and the angle above the horizontal of the ball when it was kicked.

    Could you also explain when do I use the different formulae of Range or horizontal distance
    (2 votes)
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Video transcript

What I want to do in this video is answer an age-old question, or at least an interesting question to me. And the question is, let's say I have a ledge here-- I have a ledge or cliff, or maybe this is a building of some kind. And let's say it has height, h. So let's say it has a height of h, right over here. And what I'm curious about is if I were to either-- Let's say that this is me over here, so this is me. If I were to either jump, myself-- that's not recommended for very large h's. Or If I were to throw something, maybe a rock off of this ledge, how fast would either myself or that rock be going right before it hits the ground? And like all of the other videos we're doing on projectile motion right now, we're going to ignore air resistance. And for small h's and for small velocities, that's actually reasonable. Or if the object is very aerodynamic and is kind of dense, then the air resistance will matter less. If it's me kind of belly flopping from a high altitude, then the air resistance will start to matter a lot. But for the sake of simplicity, we're going to assume no air. Or we're not going to take into effect the effects of air resistance. Or we could assume that we're doing this on an Earth-like planet that has no atmosphere. However you want to do it. So let's just think about the problem. And just so you know, some of you might say, that's not realistic. But this actually would be realistic for a small h. If you were to jump off of the roof of a one-story building, air resistance will not be a major component in determining your speed if. It was to be a much larger building, then all of a sudden it matters. And I don't recommend you do any of these things. Those are all very dangerous things. Much better to do it with a rock. So that's actually the example we're going to be considering. So let's just think about this a little bit. We want to figure out-- So at the top, right when the thing gets dropped, right when the rock gets dropped, you have an initial velocity of 0. And once again, we're going to use the convention here that positive velocity means upwards, or a positive vector means up, a negative vector means down. So we're going to have an initial velocity over here of 0. And then at the bottom we're going to have some final velocity here that is going to be a negative number. So it's going to have some negative value over here. So this is going to be negative. This is going to be a negative number right over there. And we know that the acceleration of gravity for an object on free fall, an object in free fall near the surface of the earth. We know it, and we're going to assume that it's constant. So our constant acceleration is going to be negative 9.8 meters per second squared. So what we're going to do is given an h, and given that their initial velocity is 0 and that our acceleration is negative 9.8 meters per squared, we want to figure out what our final velocity is going to be right before we hit the ground. We're going to assume that this h is given in meters, right over here. And we'll get an answer in meters per second for that final velocity. So let's see how we can figure it out. So we know some basic things. And the whole point of these is to really show you that you can always derive these more interesting questions from very basic things that we know. So we know that displacement is equal to average velocity times change in time. And we know that average velocity-- if we assume acceleration is constant, which we are doing-- average velocity is the final velocity plus the initial velocity over 2. And then our change in time, the amount of elapsed time that goes by-- this is our change in velocity. So elapsed time is the same thing. I write it over here-- is our change in velocity divided by our acceleration. And just to make sure you understand this, it just comes straight from the idea that acceleration-- or let me write it this way-- that change in velocity is just acceleration times time. Or I should say, acceleration times change in time. So if you divide both sides of this equation by acceleration, you get this right over here. So that is what our displacement-- Remember, I want an expression for displacement in terms of the things we know and the one thing that we want to find out. Well, for this example right over here, we know a couple of things. Well actually, let me take it step by step. We know that our initial velocity is 0. So this first expression for the example we're doing, the average velocity is going to be our final velocity divided by 2, since our initial velocity is 0. Our change in velocity is the same thing as final velocity minus initial velocity. And once again, we know that the initial velocity is 0 here. So our change in velocity is the same thing as our final velocity. So once again, this will be times. Instead of writing change in velocity here, we could just write our final velocity because we're starting at 0. Initial velocity is 0. So times our final velocity divided by our acceleration. Final velocity is the same thing as change in velocity because initial velocity was 0. And all of this is going to be our displacement. And now it looks like we have everything written in things we know. So if we multiply both sides of this expression or both sides of this equation by 2 times our acceleration on that side. And we multiply the left-hand side by-- I'll do the same colors-- 2 times our acceleration. On the left hand side, we get 2 times our acceleration times our displacement is going to be equal to, on the right hand side, the 2 cancels out with the 2, the acceleration cancels out with the acceleration-- it will be equal to the velocity, our final velocity squared. Final velocity times final velocity. And so we can just solve for final velocity here. So we know our acceleration is negative 9.8 meters per second squared. So let me write this over here. So this is negative 9.8. So we have 2 times negative 9.8-- let me just multiply that out. So that's negative 19.6 meters per second squared. And then what's our displacement going to be? What's the displacement over the course of dropping this rock off of this ledge or off of this roof? So you might be tempted to say that our displacement is h. But remember, these are vector quantities, so you want to make sure you get the direction right. From where the rock started to where it ends, what's it doing? It's going to go a distance of h, but it's going to go a distance of h downwards. And our convention is down is negative. So in this example, our displacement from when it leaves your hand to when it hits the ground, the displacement is going to be equal to negative h. It's going to travel a distance of h, but it's going to travel that distance downwards. And that's why this vector notion is very important here. Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, we take, essentially, the negative square root of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something. So let's say that the height is-- I don't know, let's say the height is 5 meters, which would be probably jumping off of a or throwing a rock off of a one-story, maybe a commercial one-story building. That's about 5 meters, would be about 15 feet. So yeah, about the roof of a commercial building, give or take. So let's turn it on. And so what do we get? If we put 5 meters in here, we get 19.6 times 5 gives us 98. So almost 100. And then, we want to take the square root of that, so it's going to be almost 10. So the square root of 98 gives us roughly 9.9. And we want the negative square root of that. in that situation, when the height is 5 meters-- So if you jump off of a one-story commercial building, right at the bottom, or if you throw a rock off that, right at the bottom, right before it hits the ground, it will have a velocity of negative 9.9 meters per second. So negative 9.9 meters per second. I'll leave it up to you, as an exercise, to figure out how fast this is either kilometers per hour or miles per hour because it's pretty fast. It's not something you's want to do. And this is just off of a one-story building. But you can usually figure this out. You could use this for, really, any height as long as we're reasonably close to the surface of the earth and you ignore the effects of air resistance. At really high heights, especially if the object is not that aerodynamic, then air resistance will start to matter a lot.