Different way to solve projectile motion components?

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Homework Help Overview

The discussion revolves around solving projectile motion equations to find the initial velocity (\(v_0\)) and launch angle (\(\Theta\)). The original poster provides specific measurements and equations related to the horizontal and vertical components of motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve for \(v_0\) and \(\Theta\) using given measurements but expresses confusion about the vertical component and the need for a maximum height value. Some participants question the necessity of certain values and suggest focusing on the relationship between the components.

Discussion Status

Participants are exploring different interpretations of the equations and values provided. Some guidance has been offered regarding the relationships between the components, but there is no explicit consensus on the correct approach or values at this stage.

Contextual Notes

The original poster has corrected a value for \(y_0\) during the discussion, indicating potential adjustments in the calculations. There is also uncertainty regarding the vertical position at the time of the measurements.

wortman
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Different way to solve projectile motion components?

Homework Statement


solve projectile motion formulas for [tex]\Theta[/tex] and [tex]v_0[/tex]
all measurements are in m, t is seconds
[tex]y_0=.005[/tex]
[tex]x_0=0[/tex]
the averages from my trials were:
[tex]x=8.775 m[/tex]
[tex]t=1.3 s[/tex]

Homework Equations


[tex]x=x_0+v_0*cos\Theta*t[/tex]
[tex]y=y_0+v_0*sin\Theta*t+.5a_yt^2[/tex]

The Attempt at a Solution


after plugging in my numbers gives:
for x
[tex]6.75=v_0*cos\Theta[/tex]
which as far as I can see still has two unknowns, where do I go from here?

I'm pretty confused about the y component because I'm not even sure what y is supposed to be. Don't I need a y(max) number first?
 
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You don't need ymax necessarily.

What is y when t=1.3s and x=8.775m?
 


Alright, so I got some more insight from my roommate and managed to boil the equations down to:
[tex]6.75=v_0Cos\Theta[/tex]
and
[tex]8.28=v_0Sin\Theta[/tex]

am I headed in the right direction by then solving the x component for [tex]v_0=Cos\Theta[/tex] and then combining it with the y formula to get something like [tex]8.28=6.75Tan\Theta[/tex]?

BTW I gave the wrong number for [tex]y_0[/tex]. It should be .0125 m. if you are actually checking my work
 


Looks like you're on the right track, though I'm not sure where [itex]8.28=v_0Sin\theta[/itex] comes from.
 


does [tex]6.37=v_0Sin\theta[/tex] sound better?
 


Since I don't know what y is, I don't know.

Welcome to PF, by the way.
 

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