Kinematic Equations in One Dimension | Physics with Professor Matt Anderson | M2-04

[@hattieflora1762]

Just discovering your videos for my first semester of physics. Very straightforward and helpful so far. Thank you for putting time into this.

[@unexpected9736]

Sir can you explain how y(t)=h-1/2gt^2

[@kailashsingh9737]

Very nice sir hi ji good morning ji

[@Qbtaumai]

Same thing repeated twice 😆 acceleration 😅😅

[@yoprofmatt]

Where? Thanks, Dr. A

[@jasreet7484]

@@yoprofmatt 5:59 was the clip on acceleration but it repeats at 8:00

[@jasreet7484]

Or 7:55

[@RamBhaktChandranarayan]

Quit confused, v=½gt If, v=h/t But, it is v=gt v=u+gt (v=u+at) Take u=0m/s And, v=gt.....

[@yoprofmatt]

Not sure if you were asking a question or not. It so, could you restate it?. Cheers, Dr. A

[@RamBhaktChandranarayan]

@@yoprofmatt yup, I'm asking that v=½gt or v=gt??

[@joeybasile545]

@@RamBhaktChandranarayan in order to get v, you need to derive the original position function we had, which was y(t) = h-0.5gt^2. Derive the position, and you get the velocity. So, just derive y(t). You get -gt = v(t).