Projectile Motion: Questions on Velocity & Height

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In projectile motion, the x-component of velocity remains constant when ignoring external forces. The equation Y = Voy * t - 0.5 * 9.81 * t^2 is applicable for calculating height at any given time, assuming the initial position is zero. When a projectile is released from a height, the initial position must be included in the equation as Y = Yinitial + Voy * t - 0.5 * 9.81 * t^2. The "Y" in the equation represents the instantaneous position in the y-component. Understanding these principles is crucial for solving projectile motion problems effectively.
soul5
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I have 2 question.


1) In projectile motion is the velocity in the x compontent constant, while above the release point?

2) Is Y = Voy * t - 0.5 * 9.81 * t^2 the equation used to find the heigth at any given time?


Thank You.
 
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1) If you ignore any forces acting on the projectile (and for basic courses and simplicity you usually do) or consider them to be zero, then yes.

2) If your initial position is zero, then yes. In a question where a projectile is dropped from a height (one that would have a "release point") typically you would use that height as the initial position.
Edit- The "Y" would be instantaneous position in the y component. The initial "Y" would be inserted on the right side of the equation. (Y = Yinitial+ Voy*t - 0.5*9.81*t^2)
 
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For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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