Kinematics Part 3: Projectile Motion

Things don't always move in one dimension, they can also move in two dimensions. And three as well, but slow down buster! Let's do two dimensions first. You know, like a cannonball. This is called projectile motion, and it's super important in physics. Isn't this getting fun?
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Pls repeat kinametics chapter one more time maybe you miss projectile motion.

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Had an hour lecture on it and left with nothing lmfaoo

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[@kennethringo9137]

Heads up: If you input -14^2 / 2 x -9.8 you will get 960.4. This is because the equation itself is ambiguous. You will need to input -14^2 / (2 x -9.8) and it will give you 10. This is for the checking comprehension part.

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hello My apologies , still professor Dave's equation in the last part is ambiguous .

[@renzorvillecaidoy1308]

hello its negative 10 , why is its positive 10 ?

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why we need to divide where infact there is no division part in the formula

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Professor Dave , you helped with the Projectile Motion unit for physics. I really did not know how to solve this question for physics but you helped with that so thank you.

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"We've been throwing a lot of rocks off of cliffs lately, so lest's throw one more" LMAO yes every physics book ever

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[@dennisdesalme]

This video really broke it down well. Thank you so much. I really like that you used the original kinematic equations, rather than canceling and deriving a new equation for max height, which I don't want to memorize.

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hi professor dave I have a question. you said to use -100 for the displacement. Do you always use -100? Thank you!

[@ProfessorDaveExplains]

No, that's the height of the cliff.

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@@ProfessorDaveExplains ok got it thank you!

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went through 2 weeks of this stuff in physics not understanding 2d projectile motion at all, but this cleared it up so much

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got the answer for check comprehension first try ,YES!

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[@lbtheman12345]

Is there anyway to solve for time without the quadratic formula. For example, change in Velocity (y-direction) = (acceleration due to gravity) * (time)

[@sethtulud7802]

how did you solve for the time? I tried it using quadratic formula but I got 19.288

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Hey Dave, could you explain why you used this specific kinematic equation at the end of the comprehension question? I used the delta x=v(i)t + 0.5 at2 equation and got a different result.. Thank you very much!

[@ProfessorDaveExplains]

can't use that one, we don't have any information regarding time!

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i have a question professor.....do angle of trajectory,,,affects the applied force? example if you want to reach vertical distance....do you need a lot force if its90degrees and lesser if its30degrees?

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@@JungGung it is kinematics . We don't study force here

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[@angelkhatetuano3467]

why did the time become 4.97 seconds???? can someone explain that to me!!!

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MY EXACT QUESTION RN AHHH

[@JenniferRokks]

he used the quadratic equation formula and got two different answers but time cant be negative so the time answer is the positive one.

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i honestly didnt get how you got the -4.10 and the 4.97 answer sir, i do not know about the quadratic equation. help

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check out my mathematics series! all the help in algebra you will ever need.

[@RocketFriday]

x = A + B + C 0 = -4.9 + 4.25 + 100 Using the Quadratic Formula: x = - B + | - √B^2 - 4 * A * C ------------------------------------ 2 * A Dunno why "Professor Dave Explains" cared more about viewership than sharing that knowledge but that's what you're missing.

[@dillon8518]

Why do I get d=40

[@dmytrolatysh9244]

-196/(-19.6)=10 You multiply 1/2 to (-9.8) thats why you got 40.

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Basically if u throw something at the sky , it will travel in a parbola like shape 2 motion takes place 1)vertical motion 2) horizontal motion Lets say u throw at 30degree from a horizontal(imaginary) with initial velocity of 2 m/s Now come something intersting into play , we know you are throwing the rock at an angle at the sky (like a diagonal) So lets break that single vector into 2 vectors Vx along X axis (along ground) ,Vy along Y axis (a straight line drawn towards the sky which is perpendicular to ground) Visualize this situation ! What happens ? The rock goes up to certain height and falls down (that height is called maximum height ) The staring point and ending point of rock is on the ground and this length is called the range Interesting thing here Vx represents horizontal motion Vy represents vertical motion When we spilt a vector into 2 perpendicular vectors ,its called resolution of vectors ( look this up on yt to clear things up) So what we do is , the vector u (intial velocity with we throw the rock) is spiltted into ucos30 ( along the X direction ) and usin30 ( along Y direction ) we take ucos(theta) as horizontal motion because (theta) is the angle made by our inital vector with the horizontal(the imaginary line which is parallel to the ground ) The horizontal motion of rock(ucos(theta)) remains constant in projectile motion because there is no disturbance or force The verticle motion goes on increasing , reaches maximum height and goes on decreasing . Thats it just focus on vertical motion which changes due to g When t=1s , Vy =usin(theta) -g*1 When t=2s , Vy =usin(theta) -g*2 Whent t = n s , Vy =usin(theta) -g*n

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He's trying to give so much information in shortcut that's very common on KZbin..you just have to replay the video 2 or 3 times that better than watching a 21 min video xD

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Play it in slower speed.

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Try to drink coffee. It improves focus

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i feel stupid :(

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[@madeleinecallan3153]

In order to not have to do the quadratic equation, you could know that vox = vo(costheta initial)t and voy=vo(sintheta initial ) -1/2gt^2 For objects that are simply released not shot or thrown with a vo other than zero, you can infer that no angle is made.

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