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November 27, 2014

November 27, 2014

Posted by **Jenny** on Monday, October 1, 2007 at 3:50pm.

v_yo_= +10m/s

v_xo_= +5m/s

a_y_ = -10m/s^2

So I used the formula: y=vo+1/2at^2

I drew the graph, and got a parabala.

Now, my question is, how do I find instantaneous vertical velocity at 6s using the graph?

Do I use the linear question for horizontal motion at all in this problem? x=vo(t)

Any help would be great. THanks

- Physics -
**drwls**, Monday, October 1, 2007 at 5:06pmTo use the graph only, you could either (1) measure the tangent to the y(t) curve at = 6 s, or (with less accuracy):

(2) divide the difference in y altitudes at t = 7 s and t = 5s by the interval, 2 s.

You could also get the exact answer by differentiating the y(t) function itself, to get Vy (t), but they don't seem to want you to do that.

You will not need the horizontal equation of motion to do this problem.

- Physics -
**Jenny**, Monday, October 1, 2007 at 5:14pmAlright, thanks. I drew a tangent line and found the slope at t=6, and got 3.75 m/s as my instantaneous velocity.

Just so I know... when do I have to take into consideration the linear equation governing horizonatal motion?

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