wach
  • wach
Quick question! An object is thrown downward with an initial speed of 17 m/s from a height of 81 m above the ground. At the same instant, a second object is propelled vertically up from ground level with a speed of 22 m/s. At what height above the ground will the two objects pass each other? The acceleration of gravity is 9.8 m/s Answer in units of m
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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wach
  • wach
I've basically made equations for each situation, added them as system of equations, gotten t = 81/39 and then plugged the time into each position equation/function to get height, but I'm doing something incorrect. :(
theEric
  • theEric
Hi! What I would do might be similar to what you did. For each, I would try to use the formula \(y=y_0+v_0\ t+\frac{1}{2}a\ t^2\), which assumes constant acceleration which is fine! The \(y\) values will be the same. So \(y=81+(-17)t+\frac{1}{2}(-9.8)t^2\\=\\y=0+22t +\frac{1}{2}(-9.8)t^2\) Now I think that \(\sf\color{blue}{subtracting}\) would be the way to go. That way the terms that are the same (\(y\) and \(\frac{1}{2}(-9.8)t^2\)) will be gone. So, after subtracting, we have \(0=81-0 +(-17)t-22t\\~~=81-17t-22t\\~~=81-39t=0\) Now solve for \(t\) \(0=81-39t\\\implies 39t=81\\\implies t=\dfrac{81}{39}\) So I got the same as you! Then, that is the time when the distances are equal, so let's find the distances at that time! Either equation will do. \(x=0+22\left(\dfrac{81}{39}\right)+\frac{1}{2}(-9.8)\left(\dfrac{81}{39}\right)^2\\\approx47.849112426035502958579881656805\) I calculated that with my Windows calculator. That is \(48\ [m]\) with proper significant figures. What did you get?
theEric
  • theEric
Ahh! It should be \(24.555621301775147928994082840237\). Or \(25\ [m]\).

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wach
  • wach
I got 56.444375, which is strange given that I did those steps / if not similar. I'll double-check my math.
theEric
  • theEric
Okay! I will use a real calculator to double check myself.
wach
  • wach
I double-checked, and your answer is actually correct.
wach
  • wach
I think I see where I went wrong, in mixing up my variables.
theEric
  • theEric
Okay! Cool, congrats! You were good all along, then? Haha :) Take care!
wach
  • wach
Thank you so much! I didn't think anyone would put in the time needed to check my work and whatnot. You're awesome. :)
theEric
  • theEric
Haha, my pleasure. Honestly, I looked at it because you were looking at it, and I worked on it because (1) I do that and (2, most importantly) you worked on it. :)
wach
  • wach
Thanks again! :)
theEric
  • theEric
You're welcome! Take care!

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