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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.
• 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.
• Don't mass have any effect on the velocity or time?? If not, why?
• 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.
• 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?