Difficult Projectile Motion Problem

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SUMMARY

The projectile motion problem involves calculating the initial velocity of a ball hit at a 45º angle from a height of 1.3 meters, which must clear a 3-meter wall located 130 meters away. The solution indicates that the initial velocity required to achieve this is 36 m/s, with horizontal and vertical components calculated as 25.45 m/s. The challenge arises from the lack of initial velocity data, complicating the breakdown of the motion into horizontal and vertical components. The discussion highlights the importance of understanding projectile motion equations and the relationship between the components of velocity.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions related to angles
  • Ability to decompose vectors into horizontal and vertical components
NEXT STEPS
  • Study the kinematic equations for projectile motion
  • Learn how to derive initial velocity from maximum height and range
  • Explore the implications of launch angle on projectile trajectories
  • Practice solving similar projectile motion problems with varying parameters
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to enhance their teaching methods in this area.

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Homework Statement


A player hits a ball 45º above the horizontal 1.3m above the ground. It clears a 3m wall 130m away. What was the initial velocity?


Homework Equations


?


The Attempt at a Solution



Normally projectile motion problems don't give me much trouble, but in this instance I'm not given the initial velocity and hence I cannot find the horizontal or vertical components of the velocity. It is 45º, so my thought was to find an equation in which the v_x=v_y, but nothing comes to mind.

The answer in the book is supposedly 36m/s, working backwards gives me a horizontal/vertical velocity of 25.45m/s. I can find the total time of the trip and the total horizontal distance (at a height of 1.3m) but from there I am trumped to find anything remotely similar for the other equation.
 
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Just had an epiphany

I think the 3m/130m height away is the vertex or maximum height

gonna try that out

edit: i don't think so... 100% stumped
 
Last edited:

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