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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?
• 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!
• 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.
• I assume what you are refering to is at about time mark . 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.
• 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 ]
• 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!
• How would the ball's flightpath change with air resistance?
• 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 :)
• Does this formula work if the object is thrown above the ground level and drops down to the ground level?
• No, it doesn't work in the case you mentioned, to the best of my understanding.
(1 vote)
• Why dosen't he use duration instead of Δt
• They're the exact same thing, and Δt is easier to use in equations, that and it's convention.
• 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?
• 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.
• 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?