Kinematics (projectile) - why is my solution wrong?

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Homework Help Overview

The discussion revolves around a kinematics problem involving projectile motion, specifically analyzing the jump of a biker from a slanted hill. The original poster seeks to determine the biker's speed as he leaves the jump, given the height and time in the air.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various equations related to projectile motion, noting the asymmetry of the situation and the need to consider both vertical and horizontal components of velocity. Some suggest alternative formulas that may require fewer assumptions.

Discussion Status

Participants are actively engaging with the problem, offering suggestions and clarifications. There is a recognition of the challenges posed by the asymmetrical nature of the jump, and some guidance has been provided regarding the use of specific equations. Multiple interpretations of the problem are being explored.

Contextual Notes

There is mention of potential confusion due to the initial and final displacements not being the same, which affects the applicability of certain equations. The original poster also notes a missing unit in their calculations.

Femme_physics
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Homework Statement



http://img29.imageshack.us/img29/4773/biker101.jpg

A biker's track is planned so when a biker jumps of a hill slanted at a 30 degrees angle and a height of 1.4m. During a race it's been determined that the biker depicted in the drawing stays in the air 1.65 seconds. Calculate the speed of the biker as he leaves point A.

The Attempt at a Solution



http://img815.imageshack.us/img815/2316/vovovo.jpg

Apparently this is the wrong figure. And oh yea-- I forgot to add meter per second.
 
Last edited by a moderator:
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well i like to work back from

y=yi+viyt-(gt2)/2

try that out noting the initial and final displacements from the diagram.

okay i just derived the equation you used. the reason its not working is due to this situation being asymmetrical (initial and final displacements not the same), whereas your equation only works for symmetrical situations.

try to work from the formula i have given, as it takes less assumptions then your one, just a suggestion
 
Last edited:
okay i just derived the equation you used. the reason its not working is due to this situation being asymmetrical (initial and final displacements not the same), whereas your equation only works for symmetrical situations.

Thanks a bunch! I'll keep working on that :)
 
also note that the velocity you attain from solving my equation is only the initial VERTICAL component. therefore you have to use another formula to find the total initial velocity

ill leave that up to you ;)
 
Femme_physics said:
I believed then that I used a formula that gives me V0 directly, not just its y component :)

Nice! :smile:

Just not neat. :P
 
lol, I noticed I ran out of paper as I was plugging in the figures in that formula so I crossed it out. What do you expect when you don't have a decent scanner to work with?!?

Though tbh this is easier, just writing it like that and taking a webcam pic. Tons faster, too. And I noticed that using a thick black marker it's pretty clear as well. :)
 
Femme_physics said:
lol, I noticed I ran out of paper as I was plugging in the figures in that formula so I crossed it out. What do you expect when you don't have a decent scanner to work with?!?

Well, there's less space now for bolting bugs! :)

Femme_physics said:
Though tbh this is easier, just writing it like that and taking a webcam pic. Tons faster, too. And I noticed that using a thick black marker it's pretty clear as well. :)

Yes, the thick black marker is very readable.
It's much better than what I would've expected from a webcam.

And the size in the posts is good too: not too small and not too big.

It's also nice to see a glimpse of you every now and then! :smile:
 
nice you got it

ill say once again that, in my expierience all projectile motion questions at least at my level,
can be solved by the equation i gave you and

v=vi+at (where you once again split into vertical and horizontal components)
 
  • #10
I'll keep it in mind. Thanks Lackos :)

And thanks ILS for your careful analysis of...everything :-)
 

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