Projectile Motion (Ball thrown Hortizonally)

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SUMMARY

The discussion focuses on calculating the height from which a ball is thrown horizontally at a speed of 22.47 m/s, covering a horizontal distance of 43.15 m before hitting the ground. The key equations used include Vfy = Viy - g(delta t) and yf = yi + Viy(delta t) - 0.5g(delta t)^2. The participant clarifies that the ball does not arc above the initial position but descends directly, confirming that the initial vertical velocity is 0 m/s. The solution involves determining the time of flight using horizontal distance and velocity, then applying kinematic equations to find the height.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of projectile motion principles
  • Familiarity with gravitational acceleration (g = 9.81 m/s²)
  • Basic algebra for solving equations
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  • Study the derivation of projectile motion equations
  • Learn about the effects of air resistance on projectile motion
  • Explore advanced kinematic problems involving multiple dimensions
  • Investigate real-world applications of projectile motion in sports and engineering
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This discussion is beneficial for physics students, educators teaching projectile motion concepts, and anyone interested in understanding the mechanics of horizontal projectile motion.

talaroue
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Homework Statement


A ball thrown horizontally at 22.47 m/s travels a horizontal distance of 43.15 m before hitting the ground. From what height was the ball thrown?


Homework Equations


Vfy=Viy-g(delta t)
yf=yi+Viy(delta t)-.5g(delta t)^2
yf=yi-(Vfy^2-Viy^2)/2g


The Attempt at a Solution


Since its thrown horizontally does it arc above the intial position or does it just start its decent so instead of being a upside down "u" shape it just has like a backwards "r" shape?
 
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Never heard of those shapes before. Knowing that there is no acceleration in the horizontal direction, you can get the time it takes to hit the ground with horizontal velocity and displacement.

Then you know that the initial y-velocity is 0 m/s. So just solve a kinematic equation for the height knowing the time it takes to fall to the ground with 0 initial velocity.
 
haha it was a bit unorthidox method of looking at it but I understand it now thanks.
 
No problem.
 

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