Projectile motion (part 4)
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All Comments (51)
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I am very happy to see the vidoe Solving for time when you are given the change in distance, acceleration, and initial velocity after you give this
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I Love The Video Solving for time when you are given the change in distance, acceleration, and initial velocity It Can Increase My Knowledge
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Steady I Really Like This Video Solving for time when you are given the change in distance, acceleration, and initial velocity
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I love you
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@shandytrailers dont worry I see where I went wrong
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Couldn't you do that simply but without going into quadratics?
vf^2 = vi^2 + 2ad
sqrt of vf^2 is vf
(vf + vi)/a = t
Just seems a lot simple and more acurate?
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Yea ur right, I used 26.9 not 20.9
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@p8in2008 no it was 6+20.9/2 = 26.9/2 = 13.45s..
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@thumdoctor not necessarily, it depends on which direction you chose to be positive and negative. One thing to keep in mind when choosing which direction should be postive and negative is you should choose the positive direction to be whatever the initial direction of your object is moving in. In your example you threw a penny down (initial direction) so down should be chosen as positive, which means the acceleration will also be postive.
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Time is 16.45s , he forgot to add the 6 before dividing by 2....
The people that say 13.45 is right dont know anything
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