Is There a Calculation Error in Khan Academy's Projectile Motion Video?

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SUMMARY

The discussion centers on the accuracy of the average velocity calculation in Khan Academy's video on 2-dimensional projectile motion. Users clarify that the horizontal component of velocity remains constant at 7.07 m/s until impact, contradicting the claim that it changes to zero. The average velocity formula vavg=(vinitial+vfinal)/2 is applicable only when acceleration is constant, which is not the case for horizontal motion. The conclusion emphasizes that the horizontal displacement should be calculated based on the constant horizontal velocity, leading to a displacement of 20m, not 10m.

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m_scott
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Homework Statement


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

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?
 
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You are right.
 
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.
 
m_scott said:

Homework Statement


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

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 vavg=(vinitial+vfinal)/2 if the acceleration is constant!
 
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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 vavg=(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)
 
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).
 

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