Khan academy video mistake?

[m_scott]

Homework Statement

http://www.khanacademy.org/video/2-dimensional-projectile-motion--part-3?playlist=Physics [Broken]

Skip video to 4:40. He says that the average velocity in the horizonatal direction is the same as the initial velocity (7.07m/s) because it doesn't change. but wouldn't the velocity change from 7.07m/s to 0? which means the average velocity should be 3.5m/s, right?

which means the change in displacement should equal 10m, not 20m. right?

 

[rl.bhat]

You are right.

 

[Mindscrape]

Hmm, I missed the part you are talking about, after fast forwarding, but as the other users pointed out his time was off by a factor of 2 based on when he first started.

You are wrong that the horizontal direction (did you mean to say vertical?) changes to zero though. Of course, once the ball hits the ground it may splat right in place and go to zero velocity, but we are concerned with the infinitesimal second before it hits.

 

[SammyS]

Homework Statement

http://www.khanacademy.org/video/2-dimensional-projectile-motion--part-3?playlist=Physics [Broken]

Skip video to 4:40. He says that the average velocity in the horizonatal direction is the same as the initial velocity (7.07m/s) because it doesn't change. but wouldn't the velocity change from 7.07m/s to 0? which means the average velocity should be 3.5m/s, right?

which means the change in displacement should equal 10m, not 20m. right?

The horizontal component of velocity is constant up until the moment of impact! So the person in the video is absolutely correct.

BTW: You can only be sure that v_avg=(v_initial+v_final)/2 if the acceleration is constant!

 

[m_scott]

so when you find displacement in the horizontal direction, you shouldn't ever use 0 as your final velocity? You can only use 0 as your final velocity for the y componenet?


And SammyS, he used v_avg=(v₀+v)/2 to get his answer. is the acceleration not constant?

(btw, i realize his time is wrong. i am looking at this problem based off his time)

 

[Mindscrape]

For vertical velocity it depends on where you look at the trajectory. If you want to know how high the ball goes the final vertical velocity to get to that height will be 0. If you want to know the vertical velocity before it hits the ground the vertical velocity will be the same as when it was launched (at least given the ball is launched from the ground).