IB Physics - Projectile Motion Using Energy Conservation

[@christiannemetz1203]

My man, you saved from dire confusion.

[@TheBrainFiller]

Awesome, thanks for watching!

[@learnersmixtape]

For the vertical height from the cliff, can't you just equate KEmax (on launch) to PEmax (at top of curve) and solve for the height and then add the 50m? 1/2 mv^2 = mgh (masses cancel out) and you get h = 5 m.

[@TheBrainFiller]

Yup, you’re just setting the 0 point for the potential at the top of the cliff, which is definitely allowed

[@Happy.Traveller]

How would you find the horizontal range using this shortcut method, is it possible?

[@TheBrainFiller]

Sorry for the late response…no it isn’t possible unfortunately

[@deepaksinghkanyal7573]

@@TheBrainFiller thanks for clearifying sir

[@jazminezabala7941]

Oh my gosh! Please don't stop making videos. This wasn't my exact problem but you helped me understand how to answer mine!! FYI: I had to find the final velocity of a ball that had 4kg thrown at a height of 50m at an angle of 30. BUT the ball was caught at a height of 25 meters. Makes sense that the energy at any given point should be Ei = Ef! KEWL

[@TheBrainFiller]

Sounds like a fun problem. It’s nice to hear from someone who seems to enjoy a good one. Thanks for watching!

[@mel60925]

why do you have to analayze in the y direction for the 2nd part?

[@diyaprakash3903]

The maximum height is affected by acceleration due to gravity (g) - the horizontal component (x) is unaffected by g, only the vertical component (y) is, so you have to analyse y :D

[@c.k.117]

Goated video

[@aaronalarcon2014]

for the first problem, what happened to the 30° angle you had when launching the ball? Is it not needed?

[@TheBrainFiller]

Yes exactly, it isn’t needed to find the final velocity. Thanks for watching

[@aaronalarcon2014]

@@TheBrainFiller Thanks a lot for replying! It was very helpful!