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Projectile height given time

Figuring out how high a ball gets given how long it spends in the air. Created by Sal Khan.

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  • leaf green style avatar for user khushboo Jain
    so does this mean that if I project an elephant with a velocity of 24.5 m/s, it is going to reach the same height as a feather projected with the same velocity ?
    (132 votes)
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  • leaf green style avatar for user Khushbu Patel
    why is velocity final zero??
    (29 votes)
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  • blobby green style avatar for user aneal191
    I don't see how it's possible to derive the initial velocity on the up trip from knowing the velocity on the down trip. If I shoot a bullet in the air, it's going to travel faster on the up trip than on the down trip, isn't it? On the up trip it may go 24.5m/s^2 in 1s and on the down trip 9.8m/s^2 in 2.5s. Same overall displacement, but different velocities and accelerations. Unless I'm missing something...
    (7 votes)
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    • spunky sam blue style avatar for user Stoddard Kote
      The total distance that bullet travels vertical is equal in this case to the total distance travelled up and down. The inital velocity is going to be 'slowed' down to zero (m/s) because of gravity, and the effect will be equally the same returning. Therefore the acceleration/deaccelertion is equl, the distance is equal and so will the time. This will ultimately show that the velocity at the exact same elevation (height) will be the same.

      If, the bullet hits and object higher than it was orginally fired, it will be travelling that much slower, and vice versa if the bullet travels below (say of a cliff) it will continue accelerating until impact.
      (7 votes)
  • blobby green style avatar for user Moto996
    Could anyone please confirm the displacement of 30.6 meters is only for half the time. i.e. distance reached at halfway point. or is it the total distance?
    (3 votes)
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  • piceratops sapling style avatar for user cholo.cadongonan
    So what exactly is air resistance, and why does everyone treat it as "negligible"?
    (3 votes)
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    • male robot hal style avatar for user Charles LaCour
      There are two main effects that are part of air resistance.

      One is the need for the object moving through the air to move the air out of the way as it move through it.

      The second effect is drag which is caused as the air flows around an object there is usually a low pressure region created behind the object that acts to pull back on the object. The size of this effect depends on the overall shape of the object, there is less drag on a teardrop shaped object than a box.

      Air resistance is treated as negligible in many problems because air has a low density and at speeds that are a small fraction of the speed of sound air resistance is a small effect.
      (6 votes)
  • leaf red style avatar for user Preston Tang
    At is the initial velocity zero or the final velocity? Earlier in the video he said that the initial velocity is zero.
    (4 votes)
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    • duskpin ultimate style avatar for user ck
      Even I was stuck at this but later got it:)
      Sal said that the initial velocity is zero because the ball wasn't thrown yet and it was at rest.
      Later he said that the final velocity is zero. Huh? Because when the ball was thrown up and it reached its final position, it again had to come down . Its new initial velocity was zero again because it had to come all the way down again and that was the final velocity.
      Hope that helps:)
      (3 votes)
  • blobby green style avatar for user briannanorris7
    When you are calculating displacement at the end... where do you get the equation to solve for it. I used the equation v(f)^2 = v(Initial)^2 + 2a(change in x)... However I get 3.125 meters instead... which should get the same value for the change in x as you but I didn't. What would I have to do differently to get the displacement of the object?
    (4 votes)
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    • starky sapling style avatar for user sid_07
      Well I did the calculations and for both V(F)^2 = V(0)^2 + 2ad and d = V(F) + V(0)/2 * t

      Here's what I got:
      d = V(F) + V(0)/2 * t
      d = (-24.5 + 0)/2 * 2.5
      d = -12.25 * 2.5
      d = -30.625


      V(F)^2 = V(0)^2 + 2ad
      -24.5^2 = 0^2 + 2(-9.8)d
      600.25 = (-19.6)d
      d = 600.25/-19.6
      d = -30.625

      The ball's initial velocity is the one measured at its highest point and its final velocity is the one right before it hit the ground, going downwards. In this case, the up is positive so having a negative displacement would mean it went down from its initial position.
      (1 vote)
  • purple pi purple style avatar for user vanhelsingbr
    is the velocity calculated by Average on because both the rise and fall of the object is taken into question?
    (2 votes)
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  • leafers sapling style avatar for user Prateek rana
    in the last video sal said that displacement is equal to initial velocity times time + 1/2 acceleration times time squared, but in this video an else formula was used by him, why?
    (3 votes)
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  • blobby green style avatar for user Misa Buscemi
    Is air resistance only negligible because it complicates the answer and equations?
    (2 votes)
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    • male robot hal style avatar for user Andrew M
      Yes, and in some real-life situations air resistance is small enough that we don't need to worry about it. For example if you through a heavy metal ball up in the air, you would probably not be able to measure the difference between doing it in a vacuum or in air. That's because the surface area of the ball is small compared to its mass, and it has a smooth shape, and the velocity is not very high if you are just throwing it with your hand. In some cases, the difference might be measurable but still too small for you to want to worry about.

      It is very complex to account for air resistance precisely, and that would really interfere with trying to learn the basic concepts and math involved with how objects move. You have to learn to walk before you can learn to run.
      (2 votes)

Video transcript

Let's say you and I are playing a game where we're trying to figure out how high a ball is being thrown in the air or how fast that we're throwing that ball in the air. And what we do is one of us has a ball and the other one has a stopwatch over here. So this is my best attempt. It looks more like a cat than a stopwatch, but I think you get the idea. And what we do is one of us throws the ball, and the other one times how long the ball is in the air. And then what we do is we're going to use that time in the air to figure out how fast the ball was thrown straight up, and how long it was in the air and how high it got. And there's going to be one assumption I make here-- and frankly, this is an assumption that we're going to make in all of these projectile motion-type problems-- is that air resistance is negligible. And for something like-- if this is a baseball, or something like that, that's a pretty good approximation. So we're not going to get the exact answer. And I encourage you to experiment on your own to see what air resistance does relative to your calculations. But we're going to assume for this projectile motion-- and really all of the future ones or at least in the basic physics playlist-- we're going to assume that air resistance is negligible. And what that does for us is that we can assume that the time for the ball to go up to its peak height is the same thing as the time that it takes to go down. If you look at this previous video where we plotted displacement versus time, you see after 2 seconds the ball went from being on the ground-- or I guess the thrower's hand-- all the way to its peak height. And in the next 2 seconds, it took that same amount of time to go back down to the ground, which makes sense. Whatever the initial velocity is, it takes half the time to go to 0. And it takes that same amount of time to now be accelerating in the downward direction back to that same magnitude of velocity but now in the downward direction. So let's play around with some numbers here just so we get a little bit more of a concrete sense. So let's say I throw a ball in the air. And you measure, using the stopwatch, that the ball is in the air for 5 seconds. So how do we figure out how fast I threw the ball? Well, the first thing we can do is we could say, look, if the total time in the air was 5 seconds, that means that the time-- let me write it. That means that the change in time to go up during the first half of, I guess, the ball's time in the air is going to be 2.5 seconds. Which tells us that over this 2.5 seconds, we went from our initial velocity, whatever it was, we went from our initial velocity to our final velocity, which is a velocity of 0 meters per second in the 2 and 1/2 seconds. And this isn't the graph for that example. This is the graph for the previous example where we knew the initial velocity. But in whatever that time is, you're going from your initial velocity to being stationary at the top, right? Right when the ball is stationary then it starts getting increasing velocity in the downward direction. So it takes 2.5 seconds to go from some initial velocity to 0 seconds. So we do know what the acceleration of gravity is. We know that the acceleration of gravity here-- we're assuming it's constant, although it's slightly not constant. But we're going to assume it's constant, if we're just dealing close to the surface of the earth-- is negative 9.8 meters per second squared. So let's think about it. Our change in velocity is the final velocity minus the initial velocity, which is the same thing as 0 minus the initial velocity, which is the negative of the initial velocity. And what's another way to think about change in velocity? Well, just from the definition of acceleration, change in velocity is equal to acceleration-- negative 9.8 meters per second squared-- times time, or times change in time. We're just talking about the first half of the ball's time in the air. So our change in time is 2.5 seconds-- times 2.5 seconds. So what is our change in velocity, which is also the same thing as the negative of our initial velocity? Let me get my calculator, bring it onto the screen. So it is negative 9.8 meters per second times 2.5 seconds. It gives us negative 24.5. So this gives us-- let me write it in a new color. This gives us negative 24.5 meters per second. This second cancels out with one of these seconds in the denominator, so we only have one in the denominator now, so it's meters per second. And this is the same thing as the negative initial velocity. That's the same thing as our change in velocity. And so you multiply both sides by a negative. We get our initial velocity. So that simply we were able to figure out what our velocity was. So literally you take the total time in the air, take it and divide it by 2, and then multiply that by the acceleration of gravity. And I guess you could take the absolute value of that or you take the positive version of that. And then that gives you your initial velocity. So your initial velocity here is literally 24.5 meters per second. And since it's a positive quantity, it is upwards in this example. So that's my initial velocity. So we already figured out part of this game, the initial velocity that I threw it upwards. And that's also going to be-- we're going to have the same magnitude of velocity when the ball's about to hit the ground, although it's going to be in the other direction. So what is the distance-- or let me make it clear. What is the displacement of the ball from its lowest point-- right when it leaves your hand-- all the way to the peak? Well, we just have to remember-- and once again, all this comes from very straightforward ideas, change in velocity is equal to acceleration times change in time. And then the other simple idea is that displacement is equal to average velocity times change in time. Now, what is our average velocity? Our average velocity is your initial velocity plus your final velocity divided by 2, if we assume acceleration is constant. So it's literally just the arithmetic mean of your initial and final velocities. So what is that? That's going to be 24.5 meters per second plus-- what's our final velocity? In this situation, remember, we're just going over the first 2.5 seconds. So our final velocity is once again 0 meters per second. We're just talking about when we get to this point right over here. So our final velocity is just 0 meters per second. And we're just going to divide that by 2. This will give us the average velocity. And then we want to multiply that times 2.5 seconds. So we get-- this part right over here-- 24.5 divided by 2. We can ignore the 0. That still is 24.5 That gives us 12.25 times 2.5. And remember, this right over here is in seconds. Let me write the units down. So this is 12.25 meters per second times 2.5 seconds. And just to remind ourselves, we're calculating the displacement over the first 2 and 1/2 seconds. So this gives us-- I'll get the calculator out once again. We have 12.25 times 2.5 seconds gives us 30.625. So this gives us-- so our displacement is 30.625 meters-- these seconds cancel out-- meters. This is actually a ton. This is roughly, give or take, about 90 feet thrown in the air. So this would be like a nine-story building. And I, frankly, do not have the arm for that. But if someone is able to throw the ball for 5 seconds in the air, they have thrown it 30 meters in the air. Well, hopefully you found that entertaining. In the next video, I'll generalize this. Maybe we can get a little bit of a formula so maybe you can generalize it. So regardless of the measurement of time, you can get the displacement in the air. Or even better, try to derive it yourself and we'll see how at least I tackle it in the next video.