Where does this kinematics equation come from?

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Discussion Overview

The discussion revolves around the derivation and understanding of a kinematics equation related to projectile motion, specifically focusing on the relationship between gravitational acceleration, time, initial horizontal velocity, and the angle of motion.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant presents the equation g(t)/vox=tan(theta) and seeks clarification on its origin and components.
  • Another participant suggests that the equation represents the angle below the horizontal for a projectile moving with an initial horizontal speed, explaining that it derives from the relationship tanθ = Vy/Vx, where Vy is the vertical component influenced by gravity.
  • A third participant provides an alternative formulation involving the sine and cosine components of velocity, reinforcing the connection between the angle, gravitational acceleration, and initial horizontal velocity.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of the equation and its components, but there are multiple perspectives on its derivation and application.

Contextual Notes

The discussion does not resolve the derivation of the equation fully, and assumptions about the conditions of projectile motion are not explicitly stated.

mybrohshi5
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g(t)/vox=tan(theta)

g= acceleration due to gravity (9.8m/s^2)
t= time
vox = initial velocity in the x direction

Where does this equation come from?? someone told me to use this to help me solve one of my homework problems and it worked but i have never seen it before and am having a hard time to figure out where the components of the equation actually come from?

could anyone help explain this equation to me?

thank you
 
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mybrohshi5 said:
g(t)/vox=tan(theta)

g= acceleration due to gravity (9.8m/s^2)
t= time
vox = initial velocity in the x direction
This looks like a formula for calculating the angle below the horizontal that a projectile will be moving after time t after being released with a purely horizontal speed of V0x. It's just tanθ = Vy/Vx. Vy = at = gt.
 
Hi mybrohshi5! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)

If at time t, the angle is θ and the speed is v, then

vsinθ = gt

vcosθ = vx0,

so tanθ = gt/vx0 :wink:
 
thank you both! that makes complete sense now.
 

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