Physics 3.5.4a - Projectile Practice Problem 1

[Star123Euro]

@derekowens why when we take the horizontal the acceleration is zero? please tell me

[derekowens]

@Star123Euro In this problem, the object is a projectile, which means its motion influenced by gravity only. And gravity pulls straight down. The force of gravity does not have any horizontal component, so the horizontal acceleration is zero as long as it is in free flight.

[MrNesala]

Thank you so much for these uploads. 

[TheUnchainedBeast]

I <3 you man, no homo.

[804YankeeFan]

Does the projectile objects have the inertia of the rotation speed of the earth?

[derekowens]

@804YankeeFan Yes, that's basically correct. When I throw a ball into the air, the rotation of the earth does not cause the ball to be "left behind".

[804YankeeFan]

@derekowens So, I have a last question I was wandering about. So are satellites orbiting the sun with the Earth while they are orbiting the Earth? Does this cause the Earth to be an inertial reference frame when calculating its orbit?

[804YankeeFan]

@derekowens I was wandering If I drop a ball in the air, the air's bouyancy force is acting on me and accelerating me with the weight. Am I right? So why dont people consider the extra bouyancy force?

[derekowens]

@804YankeeFan The buoyant force is very tiny, and is generally considered small enough to ignore. One could, though, include both the buoyant force and the air resistance in the calculations. The calculations get beyond the scope of this course, though.

[SohrabR93]

thanks man,grate explanation,but i don't know why in some cases other lecturers put gravity as -0.98 but it wasn't negative here!

[derekowens]

@SohrabR93 It depends on how the problem is set up. Typically a problem can be set up with up being the positive direction (in which case gravity is -9.8), or with down being positive (in which case gravity is +9.8). It can be done either way, as long as you pick one way and stick with it consistently through the whole problem.

[HairtUB]

@derekowens you say t = sqrt(2y/a) right? But if we take a = -9.8, then aren't we taking the square root of a negative number? What would you do in this case?

[derekowens]

@HairtUB It depends on how the problem is set up. If it is set up consistently, then y and a will both have the same sign in that equation, and there would be no negative square root.

[bookworm987654]

@derekowens what program do you use to write the problems?

[Varia72]

lovin the accent ;)

[lidyaFACE]

1) how come there's no horizontal acceleration? 2) the way my teacher taught it, always use -9.8 for acceleration when an object is moving downwards because the object eventually slows down. hypothetically speaking, if you used -9.8, you would get the same number for the time but just negative. can I assume both ways are right but just make sure the number is positive because time can't be a negative number? lol

[derekowens]

@lidyaFACE There are usually two (or more) ways to set up a problem. The acc. can be either positive or negative, depending on how it is set up. For a projectile, though, the horizontal and vertical motions are always independent, and the acceleration of gravity only applies to the vertical motion. For a projectile, the horizontal acceleration will be zero.

[MsBiebaholic]

Ummm, what we learned is school is that the only equation we can use for the horizontal component of the velocity is " x = vot + xo " because it is executing Uniform Rectilinear Motion ( We can't use " x = xo + vot + 1/2 a t^2 " because that is an equation for bodies executing Uniformly accelerated rectilinear motion ) ..... ? ----- and is there an easy way to learn the equations of projectile motion because we're not given enough time to prove them during the exams ..

[derekowens]

@MsBiebaholic Yes, that is correct. The two equations you mention are actually the same, since horizontally there is no acceleration so the 1/2 a t^2 term reduces to zero. The larger equation simply reduces to the smaller in this case. I don't know an any easy way to memorize the equations of motion, but even if it's just by brute force or practicing, memorizing them is certainly a good idea.

[rykneeto]

I have a quiz on this in 2 hours...helping immensely :)

[omar3211]

how did u get acclearation is 9.8 from nowhere

[derekowens]

@omar3211 Acceleration near the earth's surface, due to gravity, is 9.8 m/s^2 (or very close to that). That's basically a constant (if you're on earth).

[McBeastin1]

can you teach my physics class lol

[1001Falcon]

owensacademy?



[9448234580]

very nice

[Ell4Sh]

im really bad at physics im watching this to get a better understanding obviously. i would make the top of the cliff 100m and the bottom 0m.. id also make my a -9.8 m/s.. i guess what im asking is how do you lay it out that way?

[derekowens]

@Ell4Sh Yes, that approach would also work. There are typically multiple ways to set up a problem, all of which should give the same final answer. I set it up with 0 a the top because all of the motion and the acceleration are downward, in this problem, and setting it up this way makes all the displacement, velocity, and acceleration numbers positive. The other approach is fine, though.

[LEMcC15]

i love this. you're literally the man

[DeepSpace8andahalf]

Question: WHY should an aircraft be observed traveling as fast while penetrating a wall as it did flying through AIR? ... whats up with that?

[anoorcc]

@ derekowens; When finding the initial horizontal velocity, should time be doubled @ 7:21 so that it account for the whole horizontal time. So instead of 4.52 should it be 9.04. Can anybody else see what I'm talking about as well?

[derekowens]

@anoorcc Not in this case. If it were a symmetrical parabola, going up and then back down, then we could double the t to find the whole time. In this case, though, we only have half of such a parabola.

[wanatchaphorn]

Hi ,derekowens I come form Thailand. I happy so much. you give knowledge of physics for me I think that you are kind  person.

I thank you ps.i'm sorry for bad english

[emmalainesmiles]

What if a value's given in the Vo at 3:34? How will you find the t then?

[lonogugeadult]

@emmalainesmiles you'd use the equation y = yo + vo(t) + 1/2a(t^2) and plug in what you know for the vertical component to solve for t



[earlcamblelover]

What about when horizontal distance is not given ??

[superbaby0907]

love the fact that its colourful!

helps to clearify things a lot!

[gambitx24]

you are very good.

[r0oTsWe]

oh man...you do good job ,

thank..I like & SUB

:)

[gareler]

13337 views... ELITE

[rassinlababidi]

cant you use v=at+v0?

[darkflamehaze666]

UNDERstood! thanks!

[AVerbene]

thanks to you, i am going to pass my physics midterm tomorrow

[ClarinetgirlM]

OMG THANK YOUUUUUU

[juschecknin]

Thank you! You helped me so much, but I still have a question.

Why is there no horizontal acceleration if the rock is being thrown forward horizontally?

[derekowens]

@juschecknin That's because we are just considering the flight of the rock from the moment right after release to the moment right before it lands. There is acceleration during the act of throwing it, but that would be a separate little physics problem. Once it is released, it simply coasts horizontally at a constant velocity. During the flight from release to impact, the horizontal velocity remains constant (assuming we ignore air resistance).

[juschecknin]

@derekowens Ok, I see now. Thank you so much!

[Palmar3s]

@juschecknin

because gravity doesn't affect it because it's going horizontal, that is why there is no horizontal acceleration, but if it's vertical then yes, because gravity pulls down, gravity doesn't work from side to side (horizontal) therefore there is no acceleration. As far I have seen, every time acceleration is mentioned I think of gravity working on the vertical axis, pulling down, never it acts sideways, unless there is another force of acceleration. I think it goes like that :/

[AlexC1210]

sir, you have saved me.

i'm so happy right now i want to cry.

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[ClarinetgirlM]

@AlexC1210 LOL i feel you

[jojosh234]

Why does your gravity at positive? it should be at negative because it's going to negative Y axis

[jojosh234]

@jojosh234

the vertical distance should be at negative because it's going down to Y axis,

[derekowens]

@jojosh234 You can set it up either way. If you think of up as the positive direction, then gravity is negative. If you think of down as the positive direction, then gravity is positive. Either way is okay, just don't switch it in the middle of a problem.

[avgrech]

10x a million ma 2 morro i got a test aout thease and i was sick all week no idea what to do but now i got a clue 10x m* ur da best vry good explenation btw

[jioddon]

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[Denseworldproduction]

thanks WOW! your 8 min tutorial rlly helped me do a problem that wasnt even the same as this, so basically you improved my understanding :)

[RuthBuzzzzz]

i do not understand why the acceleration is positive in this problem! i thought because of gravity pushing down 9.8 would always be negative.

[rockster16]

@RuthBuzzzzz It would be negative, but when you square anything its always the absolute value of that, so the negative or positive wouldn't matter

[VaisaNacia]

Thank you so much!

[alkhor999]

what is Y?

[derekowens]

@alkhor999 I typically use x to indicate the horizontal position and y to indicate the vertical position.

[yo3443]

@derekowens you're videos are awesome! Did you get a perfect 50 for physics? I'm sure you did. You really know you're stuff.

[Applecheria]

I am at the verge of failing Physics class. Chemistry was a mess and I do not want to Jeopardize my GPA! Thank you so much for your help!

[jdenise4life]

You explain this better than the book. better than my physics teacher.

Thank You so Much. This is is such a help. Thank YOU!

:D

[Phleege]

thank you this really helped, mabye ill actually be able to pass my physics ib exam

[colorgreens]

your hand writing is beautiful! :)

[021iran1]

dude u rock :P THANK YOU!!!!!!!!!!

[Satanic1992]

just a bit of confusion you use v as your initial velocity, but what do you use for your final velocity, like when the object hits the ground. I normally use u for initial and v for final or it doesn't matter?

[derekowens]

I typically use v_0 for the initial velocity. That's v with a zero in the subscript position. It basically means "velocity at time zero" or "velocity when t=0". Then I use v by itself for velocity at some later time. Using u works also, though. Good notation definitely helps, but it's not as important as the concept.

[TheCarlosfgm91]

Awesome! thanks for the help! please keep on making physics videos, they're a ton of help! 5stars!

[oPOPNLOCKo]

Can you PLEASE fire my physics teacher and replace him. He is sooooooooooooooooooooooooo bad at teaching and my grade suffers from it. Sir, you explained this so well. the reason im watching this is because im taking his mid term tomorow and i need to review.

This helped SO much you dont understand. I honestly wish you were my physics teaching and you should be proud of that because MR ECKLE (my teacher) SUCKS!!!!! but really thanks again. im watchin all the practice vids.