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arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
let me draw it out
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
Ah @eashmore is here. He eats these problems like popcorn at a movie.
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.1
Only if the butter isn't too thick!
 2 years ago

arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
its not a problem, its just one stupid question, i am not able to draw properly with this editor lol
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
Draw a picture, take a picture of it and upload it. I do that all the time. Let us know when you're ready
 2 years ago

arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
yep! one second
 2 years ago

arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
My question is, will the time from O to P will be equal to P to S when there is air drag?
 2 years ago

arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
Popcorn! lol
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
No, it is not. Classic question in kinematics. And why not?
 2 years ago

arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
But the trajectory is different from P to S right? so why not?
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
On the way up to the top of it's trajectory, there are two forces acting: gravity and air resistance. Air resistance of course is always in the direction of v, the velocity vector. Air resistance has negative y component and gravity is in the negative y direction. On the way down, the air resistance has positive y component and hence the net force in the y direction is less than on the way up. The result is the magnitude of the force on the object on the way up is more than it is on the way down. So the time required to reach the top of its trajectory is created than the time required for it to travel from the top to the bottom.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
To convince yourself of this, take a large piece of paper, like a newspaper. Crumble it up into a rough ball. Throw it up and notice it takes less time to reach the top of its trajectory than it does to come down.
 2 years ago

arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
Nah, i understood it from, when i serve in tennis, it takes more time to go up but land a bit faster in the box! Gravity and air restistance! I always liked the symmetry for projectiles until yesterday i watched this lecture on air drag! Thanks a lot @JamesJ !!!!
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.1
This effect is most noticeable on the smaller curves.
 2 years ago

arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
/offtopic Lets make it 1000 medals tonight @JamesJ ! :D
 2 years ago

arcticf0x Group TitleBest ResponseYou've already chosen the best response.0
haha leave it if you dont care :P
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
Ah! I haven't looked at that number for a long time. Yes, why not? ;)
 2 years ago
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