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Deriving max projectile displacement given time
Deriving a formula for maximum projectile displacement as a function of elapsed time. Created by Sal Khan.
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- How do we know that the total time up is exactly half of the total time in the air?(28 votes)
- You can also prove it algebraically. Consider a initial speed V_o, at the highest point the velocity V = 0. Then using V = V_o + a*t(max height), we plug in a = -g and we see that 0 = V_o - gt or V_o/g = t(max height). Now lets consider total trip. We use Y = V_o*t + 1/2*a*t^2; Y = 0 here, because we are back down on the ground - > 0 = V_o*t - 1/2*g*t^2. So 0 = t(V_o -1/2*g*t). We get t = 0 or t = 2V_o/g. The first solution is at initial time it is on the ground. The second t is when it comes back down. We see that t(max height) = V_o/g and t = 2V_o/g so it is half the time. Hope this helps!(16 votes)
- Why isn't delta t squared divided by 4 as well? Sal just divides 4.9 by 4. The whole numerator should be divided by 4.(21 votes)
- I assume what you are refering to is at about time mark5:25. You can look at the s=(4.9 * v^2)/4 as s=1/4 * 4.9 * t^2 and both the 1/4 and the 4.9 are just numerical constants being multiplied with t^2 so you can combine them.
If you step back a step where he has s = ((4.9 * t)/2) * (t / 2) you can rewrite it as (1/2 * 4.9 * t) * (1/2 * t). Using the associtivge proerty of multiplication you can rewrite this as s = 1/2 * 1/2 * 4.9 * t * t. The 1/2 * 1/2 * 4.9 becomes the 1.225 and the t * t becomes t^2 givong you s = 1.225 * t^2.
I think you are confusing how to simplify (a + b) / c with (a * b) / c. (a + b) / c is equal to a/c + b/c where as (a * b) / c = a/c *b or a * b/c or a * b * 1/c.(39 votes)
- I was kinda puzzled when he changed 9.8 m/s^2 to 4.9 m/s^2. I know he just divided 9.8 by 2 but didn't understand why. So, while thinking about this problem I came up with another formula than the one Sal got at the end.
My question is, whether this formula is valid:
[ (9.8 * ▲T)/4 * ▲T/2 ](9 votes)
- The entire equation he was working with was: DeltaV = -9.81 * DeltaT / 2. The first simplification he did was just to divide the -9.81 by the 2 so we then have DeltaV = -4.9 * DeltaT.
Your factor looks like it will simplify to (9.8 * DeltaT^2) / 8 which is 1.225 * DeltaT^2! Good Job!(8 votes)
- How would the ball's flightpath change with air resistance?(6 votes)
- If air resistance was included, the ball would experience drag, and the drag would produce a slower time. The greater the height or the more massive an object, the more the air resistance matters!
Hope that helps :)(10 votes)
- Why dosen't he use duration instead of Δt(4 votes)
- They're the exact same thing, and Δt is easier to use in equations, that and it's convention.(10 votes)
- Does this formula work if the object is thrown above the ground level and drops down to the ground level?(8 votes)
- No, it doesn't work in the case you mentioned, to the best of my understanding.(1 vote)
- Don't mass have any effect on the velocity or time?? If not, why?(3 votes)
- no, this is a common misconception. eventually it is the acceleration which matters, not the mass.
if a 1 ton brick, and a 1 kg block are dropped from the same height, neglecting air resistance, they would land at the same time.(8 votes)
- What if we projected a pistol bullet and cannon ball at same time, same height, same angle, also neglecting air resistance?
Will they both land on the ground at the same time?(5 votes)
- yes! Projectile motion is independent of mass of the object involved! Moreover all objects which have the influence of acceleration due to gravity (g), will move in the same path! this is because 'g' does not depend on the mass of the object; g= GM/R^2, where G is gravitational constant, M is mass of earth and R is its Radius.(5 votes)
- This entire process is based on the assumption that you have a specific time. How could you solve if you didn't have the time?(4 votes)
- You could make a table for the different times but in the end you won't have a set answer.(5 votes)
- This is assuming the ball is launched directly from the ground. If I throw the ball from my hand, about a meter up, and we stop the stopwatch when the ball hits the ground, we can't assume it peaks at t/2. How would we go about it in this case?(4 votes)
Just want to follow up on the last video, where we threw balls in the air, and saw how long they stayed up in the air. And we used that to figure out how fast we initially threw the ball and how high they went in the air. And in the last video, we did it with specific numbers. In this video, I just want to see if we can derive some interesting formulas so that we can do the computations really fast in our brains, while we're playing this game out on some type of a field, and we don't necessarily have any paper around. So let's say that the ball is in the air for delta t. Delta t is equal to time in the air. Then we know that the time up is going to be half that, which is the same thing as the time down. The time up is going to be equal to delta t-- I'm going to do that in the same color-- is going to be equal to the time in the air divided by 2. So what was our initial velocity? Well, all we have to do is remind ourselves that the change in velocity, which is the same thing as the final velocity minus the initial velocity. So the final velocity-- remember, we're just talking about half of the path of this ball. So the time that it gets released, and it's going at kind of its maximum upward velocity, and it goes slower, and slower, slower, all the way until it's stationary for just a moment, and then it starts going down again. Now remember, the acceleration is constant downwards this entire time. So what is the final velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0 minus our initial velocity, when it was taking off. That's our change in velocity. This is our change in velocity, is going to be equal to the acceleration of gravity, negative 9.8 meters per second squared-- or the acceleration due to gravity when an object is in free fall, to be technically correct-- times the time that we are going up. So times delta t up, which is the same thing. I won't even write it. Delta t up is the same thing as our total time in the air divided by 2. And so we get negative the initial velocity is equal to-- this thing, when you divide it by 2, is going to be 4.9 meters per second squared-- we still have our negative out front-- times our delta t. Remember, this is our total time in the air, not just the time up. This is our total time in the air. And then we multiply both sides times a negative. We get our initial velocity is just going to be equal to 4.9 meters per second squared times the total time that we are in the air. Or you could say it's going to be 9.8 meters per second squared times half of the time that we're in the air. Either of those would get you the same calculation. So let's figure out our total distance, or the distance that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air-- so it's final velocity is 0. We're talking when it gets to this peak point, right over here-- that's from two videos ago-- that peak point right over there. So our average velocity is just going to be this stuff divided by 2. So it's going to be 4.9 meters per second squared times delta t over 2. So this right here, this is our average velocity. Velocity average. So let's stick that back over here. So our maximum displacement is going to be our average velocity-- so that is 4.9 meters per second squared-- times delta t, all of that over 2. And then we multiply it again times the time up. So times delta t over 2 again. This is the same thing. These are the same thing. And then we can simplify it. Our maximum displacement is equal to 4.9 meters per second squared times delta t squared, all of that over 4. And then we can just divide 4.9 divided by 4. 4.9 divided by 4 is-- let me just get the calculator out. I don't want to do that in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum displacement is going to be 1.225 times our total time in the air squared, which is a pretty straightforward calculation. So this is our max displacement, kind of how high we get displaced. Right when that ball is stationary, or has no net velocity, just for a moment, and starts decelerating downwards. So we can use that. If a ball is in the air for 5 seconds-- we can verify our computation from the last video-- our maximum displacement, 1.225, times 5 squared, which is 25, will give us 30.625. That's what we got in the last video. If the ball's in the air for, I don't know, 2.3 seconds-- so it's 1.225 times 2.3 squared-- then that means it went 6.48 meters in the air. So anyway, I just wanted to give you a simple expression that gives you the maximum displacement from the ground, assuming air resistance is negligible, as a function of the total time in the air. I don't know. I find that pretty fun. And it's a neat game to play.