Calculating Vertical Velocity from Projectile Motion Data

AI Thread Summary
The discussion revolves around calculating vertical velocity from projectile motion data obtained during a lab experiment involving water balloons. The initial height was determined to be 9.2 meters based on an average fall time of 1.37 seconds, using the equation y = 1/2gt^2. Participants debated the correct kinematic equation to find vertical velocity, with confusion arising over whether to use initial or final velocity. The experiment involved throwing the balloons upwards, which required a different approach than simply dropping them. Clarifications were made regarding the appropriate time to use in calculations, emphasizing the need for accurate input values to determine the correct vertical velocity.
Dejahboi
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Homework Statement



Alright we had a lab and we threw water balloons off the college's stadium to determine the height.In result the I got 9.2m from an average of 1.37sec using y=1/2gt^2

Okay, this is where i get confused to use the right equation to determine vertical velocity.

Homework Equations



y=Yo+Vot+1/2at^2

The Attempt at a Solution



I did v=u+(9.8m/s^2)(2.81) to find the velocity. Is that right?
 
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Hi Dejahbol, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif
Dejahboi said:
Alright we had a lab and we threw water balloons off the college's stadium to determine the height.In result the I got 9.2m from an average of 1.37sec using y=1/2gt^2
You threw them? Or dropped them? Is 9.2m about right?
Okay, this is where i get confused to use the right equation to determine vertical velocity. [...]

I did v=u+(9.8m/s^2)(2.81) to find the velocity. Is that right?
I'm confused, too, because it's not clear whether you are trying to find u or v. Which?
 
Last edited by a moderator:
Sorry i wasn't very elaborate on my information. The estimate height dropping the balloon at rest is 9.2 m. That was based on the average time of 1.37 seconds. For the second part of the lab we were given the choice to throw them down or up, and we decided with our group to throw it up and took 2.81 seconds to hit the ground. The question asks "Using the height of the building, find the vertical velocity with which you threw the balloon."

And thanks for the welcome :)edit: I wasn't sure which kinematic equation to use for this particular part of lab.
 
Dejahboi said:
For the second part of the lab we were given the choice to throw them down or up, and we decided with our group to throw it up and took 2.81 seconds to hit the ground. The question asks "Using the height of the building, find the vertical velocity with which you threw the balloon."
You need an equation involving u as the only unknown, e.g., s = ut + ½at²

(You'd be seeking initial velocity, u)
 
So I end up doing this:

9.2m-(1/2)(9.8m/s^2)(1.37s)2
/(1.37s)

= 2.3x10^-3... i think i did something wrong lol.
 
I've drawn it out for you below.

Granted you threw the balloons straight forward as Vo and let them fall rather than launch them up and then down which would be an entirely different question. I've also solved it based on the premise that the information you gave in the OP is accurate so take a look.

http://i48.tinypic.com/2wp8nds.jpg
 
Wait i need vertical velocity :/. I got that part down and thanks for helping ;)
 
Dejahboi said:
So I end up doing this:

9.2m-(1/2)(9.8m/s^2)(1.37s)2
/(1.37s)

= 2.3x10^-3... i think i did something wrong lol.
If it spent 2.81 secs in the air, why are you using 1.37 in the formula? :confused:
 
didn't he say it spent 1.37 in the air?

a bit confused now
 
  • #10
Thanks. I plugged in the wrong values... *sigh* 2.31. Sorry it's like 2 am and I've been up all day.
 
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