How far above the release point will the ball and stone pass each other?

[meggy8716]

I know I should be able to do this problem but for some reason it's not clicking with me...please help!

Homework Statement

a ball is thrown vertically upward with an initial speed of 19 m/s. Then, .97 s later, a stone is thrown straight up (from the same inital height as the ball) with an inital speed of 31 m/s. How far above the release point will the ball and stone pass each other. Answer is units of m. Gravity is -9.8 m/s^2.

Homework Equations

I tried using v^2=v0^2*2ax and x=x0+v0t+0.5at^2

The Attempt at a Solution

I solved 19(.97)+0.5(-9.8)(.97)^2 and got 13.82 m for the ball after .97 s.

After this I got lost and didn't know what to do. I know this means that when the ball is at 13.82 m/s that the stone is just being thrown. Please help!

 

[Mindscrape]

Do you have a graphing calculator? If so that might be a handy visualization tool.

You are going to have two kinematic equations with distance as a function of time, and you want where they intersect. You will will probably end up with two answers, which means that one won't make sense and should be thrown out.

 

[meggy8716]

I used to have one but I have no clue where it is at anymore. We aren't aloud to have them in this class so I have kinda lost track of it while I was moving.

 

[Mindscrape]

Well, your first equation looks okay if you put the time parameter back in.

What does the second one look like?

 

[meggy8716]

I actually figured it out today right after class (guess it got me in the right type of mode to do it). Sometimes I just don't look at them the way I should be. But what I ended up doing is plugging in 19 m/s for v, -9.8 for g (a) and 0.97 in for t into the equation
v = v0 + at
This gave me the answer of the velocity being 9.494 m/s when the ball is at 0.97 s and 13.82 m.
Next, I plugged these numbers into the eqation
x = x0+ v0t + 0.5at^2
and set this equal to the other side. this ended up looking like
13.82 + 9.494t + 0 = 0+ 31t+ 0
solving for t I got 0.643 s.

finally, I plugged all these numbers I solved for into
x = x0 + v0t +0.5at^2
this looked like
13.82 + 9.494 (0.643) + 0.5 (-9.8) (0.643)^2
The final answer ended up being 17.9 m.

=)