Question

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

Answer 1

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. :(

Answer 2

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?

Answer 3

Ahh! It should be \(24.555621301775147928994082840237\). Or \(25\ [m]\).

Answer 4

I got 56.444375, which is strange given that I did those steps / if not similar. I'll double-check my math.

Answer 5

Okay! I will use a real calculator to double check myself.

Answer 6

I double-checked, and your answer is actually correct.

Answer 7

I think I see where I went wrong, in mixing up my variables.

Answer 8

Okay! Cool, congrats! You were good all along, then? Haha :) Take care!

Answer 9

Thank you so much! I didn't think anyone would put in the time needed to check my work and whatnot. You're awesome. :)

Answer 10

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. :)

Answer 11

Thanks again! :)

Answer 12

You're welcome! Take care!