2-dimensional projectile motion (part 3)

[Kricardose]

I am very happy to see the vidoe Completing our first example from parts 1 and 2 after you give this

[bebeheuy]

I Love The Video It Can Increase My Knowledge Completing our first example from parts 1 and 2

[Mjhond]

Steady I Really Like This Video Completing our first example from parts 1 and 2

[bladchou]

Thank you, thank you, thank you.

[Yasoofeus]

Wow! Why do they complicate things so much in the textbooks?

[treeandplant]

@Yasoofeus Because those cocksuckers want to showoff.

[tkatiqah]

you waste a lot of time drawing the things. i mean, can't you just use a marker and paper? idk, something else easier to use.. other than that your videos are awesome! thanks a lot

[Muhsin168]

Sal..there is a big mistake..on the previous video ...and u took the same value in this video too

the total time should be "sqrt2"

not "2 sqrt2///ok

[Muhsin168]

Sal..there is a big mistake..on the previous video ...and u took the same value in this video too

the total time should be "sqrt2"

not "2 sqrt2///ok

[dmk234]

No guys t = sqrt(2)/2 = .707 sec

which doesn't half things, but divides them by a quarter.

[Nejitheultimate]

thank God for selfless people like you, Sal.

[ecophysicsEMF]

Yes to comments saying that he dropped a factor of dividing by 2. Total time in air should be square root of 2 sec, not two times square root of 2 sec!

Videos as a whole are fabulous.

[ecophysicsEMF]

Yes to comments saying that he dropped a factor of dividing by 2. Total time in air should be square root of 2 sec, not two times square root of 2 sec!

[chaosavenger2]

lol you dont know how to do math son

[V1CanisMajoris]

This is realy helping me to understand more, although i understood before, i do alot more now, TY : )

[endysea]

Again (as in previous part), you made the mistake of dividing 10 by 5root2 to get t=root2. It should be t = 5root2/10 = (root2)/2

[Bella329]

for finding how high the ball went: why are you using the average velocity?

[naruto2710]

total time= 1.414

range= horizontal velocity*total time flight

7.07*1.414= 9.99698m =10m

[lhud2020]

You are awesome! Thanks for explaining the concepts in an understandable way

[seniorgomez666]

you rock, i have a physics test tomorrow, this is so simplified!

[gur0004]

BY him making mistakes proves that he is a normal human being just like everyone else. Do i need mention that it's not because he doesn't understand what he is doing. Anyway Mistakes are good for us as long as we learn from them.

[madk777]

War is very bad!

[madk777]

Amazing stuff! Thanks so much for the free video lessons, they are helping me immensely with my mcat prep. I have never had to take physics before (even in high school, believe it or not), and the way you explain these concepts makes them easy to grasp. I especially like that you mention and give examples of the practical applications of trig and motion. Albeit a little grim at times, they help to keep me motivated b/c the concepts are actually useful somehow. Thanks again for making math fun :D

[jackncoke99]

time to reach highest = sqr root of 2/2

time of flight = sqr root of 2

distance = 10m

[Bella329]

@jackncoke99 did you get 5/2 for the vertical distance?

[wiclepotter4eva]

oh thank god somebody else noticed the mistake! in the previous video i was the only one who noticed so i was afraid i was wrong.

[johnny19931993]

so basically...

for any 2D projectile motion

with an initial velocity v and angle θ

assuming gravity is the only force acting on it

the vertical displacement (height) is (v^2)(sin^2 θ)/g

and the horizontal displacement is 2(v^2)(sinθ)(cosθ)/g

[yamahabibi]

your no 1 teacher in the youtube .

[matty1258]

when he wrote 10 x 2 he used a . for the times. i am fine with using that when you only have letters but how do you tell it abart from a decimal point when your using numbers? is it the maker, sorry don't know your name, being messy? or do you only use fractions in physics?

[OldTreeCutter13]

sorry for posting that three times, my computer was running slow and I thought it hadn't posted it, so I kept hitting 'post'. oops...

[OldTreeCutter13]

It's just because time for the upward motion should be √2/ 2 seconds....making the total time in the air √2 seconds, all that happened is that he did manipulated the equation ∆v=at by dividing acceleration (-10 m/sˆ2) by ∆v (5√2 m/s) instead of doing the opposite [(∆v/a)=(t)]. If you re-do that (if you are working it out) then the rest of your numbers should be correct. It's not a big deal, just look at the problem and use what he has taught you in the past videos.

[OldTreeCutter13]

It's just because time for the upward motion should be √2/ 2 seconds....making the total time in the air √2 seconds, all that happened is that he did manipulated the equation ∆v=at by dividing acceleration (-10 m/sˆ2) by ∆v (5√2 m/s) instead of doing the opposite [(∆v/a)=(t)]. If you re-do that (if you are working it out) then the rest of your numbers should be correct. It's not a big deal, just look at the problem and use what he has taught you in the past videos.

[MarvelsofaLifetime]

Excellent video, and I love the intuition!

[yrollgayanth]

life saver!

[a7thd]

t = sqt 2 / 2

[scarlettbildhauer]

awesome

[bloodstone1445]

V =

10cos45 = 5.25

5 sqrt 2 = 7.07

?

[gryphonkeen]

your calc is probably on radial settings,

10cos45 = 7/07106

[gryphonkeen]

opps 7.07106

[max1916251080]

Your calc is in radians. put it in degrees and you will get 7.07

[jorgex2000]

Thank you so much. Your videos are really helpful and very easy to understand. Thanks alot

[RafiqueBrowne]

Man greetings from the Caribbean. I am a native english speaker who's doing Electronics/Telecom in SPANISH so sometimes I am double confused...thanks man !!!!!!!!!!

[x4812k]

"not that I advocate these things, war is very bad"

hahaha

[danceathletics92]

nooo ! everything is half of what he is saying it is (i know everyone already caught and mentioned his mistake) but it is kinda confusing .... SO @ the end .. the ball he shoots reaches a max height of 2.5 m not 5..and it lands 10 m to the right not 20 ! right ?

[baeronautics]

@danceathletics92 its not even 10 its 7.07106m

[alia5551]

@danceathletics92 yeah he does. But its just a human error anyways. Thanks to you as well :)

[madscirat]

Disregard my last comment, Megalnhaler is right I think.

[DoubleKx]

no he is not. the range is 20 m. horizontal velocity is uniform. There's no horizontal acceleration, so the horizontal velocity remains constant unlike vertical velocity which alters due to gravity

[madscirat]

The answer from the last problem was sqrt2/2 which makes the total distance sqrt2 (because we were only solving from ground to max point). Sub that into the above and you get sqrt2*5 which is 7.071...

[ahsimpleman]

I seen it like 5 times and when he said that V1=a*t and solve for that and you get t= square root of 2.

I don't see how he did the math on that to get square root of 2, i'm not getting that. -5square over -10 doesn't give yoiu the value of square of 2. So how did he do it

[ahsimpleman]

to my last comment :

if you 4((-5sqr2)/-10)=2.82 which is the value that he us saying that will come from 2sqr of 2.

Tell me how my math is wrong , if it is or did I miss something?

[glacialmind101]

Best to think of it in geometrical terms. We're dealing with a 45-45-90 triangle here so the ratios of the sides are c=sqrt2, b=1, 1=1. If you visualize sine and cosine as a radius of the unit circle, then at a 45 angle it intersects the circles side at point (.5,.5). .5 *10*sqrt2= 5sqrt 2. If you dont get unit circle, see Sal's video on it, its an eye opener.

[MegaInhaler]

range should be 10 m? yeh?

[ameenebrahim]

He himself confessed about his mistake in the first comment of the previous video. Anyways, he's a brilliant teacher! Everyone makes mistakes

[SimeonBerson10101]

i agree with everyone else. he's made a mistake!

[frankkkk99]

yeah he made a mistake solving for t. its sqrt of 2 / 2

[megaelliott]

time to reach the peak was equal to sqrt(2) / 2.

That means that the total time from starting at the ground, to the peak, and back down again would be twice as much so the total time is equal to 2*(sqrt(2) / 2)= sqrt(2).

[magdalenarmes]

There is nothing incorrect about the solution.

d= 5 sq root of 2 x 2 sq root of 2 = 5 x2 x sq root of 2 x sq root of 2 = 10 x 2 = 20 m

[jamleafer]

all the answers since he found time were off by a factor of incase anyone was wondering, not a biggy, cool problem haha

[mewejo]

Awesome thanks! :)

G is 9.81 m/s Sqrd not 10 m/s Sqrd :P

[megaelliott]

he was rounding. For simplicity

[ibethgc]

-5*√2=-10*t √2/2=t ??

then total time wouldn't just be √2 ?

[ibethgc]

now... im confused xD

[canadiangurl523]

I don't understand how you got square root of 2 for time. I thought you would divide 5sq.2 by 10 to find time....

[anava84]

yeah time should equal square root of 2 over 2.

[unknownlove19]

he mentioned in the previous video he did it wrong in the comments section

your doing it correctly don't worry

[quarkqueen]

Great vids.

Thanks for the help :D