Find minimal speed for ball to clear wall

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Homework Help Overview

The problem involves a golfer hitting a ball at an initial speed Vo and angle θ to determine the minimum speed required for the ball to clear a wall of height h, located a distance d away. The scenario assumes no air resistance and presents a challenge when the angle θ results in a condition where tan(θ) < h/d.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the range formula and its application to the problem, questioning the meaning of the components Vxo and Vyo. There is also a consideration of whether to solve for initial velocity using a different formula.

Discussion Status

Some participants have provided clarifications regarding the components of the initial velocity vector and the limitations of the range formula when the projectile must clear a height. There is an indication that a specific formula may be used to solve for the initial velocity, but no consensus has been reached on the best approach.

Contextual Notes

The discussion highlights the importance of understanding the conditions under which the range formula applies and the implications of the angle θ on the problem setup. There is a mention of potential pitfalls if the angle does not meet certain criteria.

leonne
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Homework Statement


Golfer hits a ball with speed Vo at angle @ above the horizontal ground. angle is fixed and no air resistance what is min speed for which the ball will clear a wall of height h and distance d away. your solution will get you in trouble if angle @ is such that tan@ < h/d explain


Homework Equations


rvac= ((2Vxo)(Vyo))/g


The Attempt at a Solution


I just have a question, after googling range formula i found R=v^2/g sin2@ now why in my book it gave me rvac= ((2Vxo)(Vyo))/g ? could not find much info on what Vxo or Vyo stand for. There was a example and I got the same answer when i used R=v^2/g sin2@ but it never showed what the Vxo equals, they just skipped all the steps.

Thanks
 
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Vxo and Vyo are the x and y components of the initial velocity vector. The range formula that you googled gives the horizontal distance (range) only when the projectile returns to the same height from which it was launched. This is not the case here because the projectile has to clear a wall of height h.
 
o ok thanks for the info
 
so would you use this formula and solve for initial velocity? y= (x(Voy)/Voc) - 1/2 g (x^2 /V^2ox)
 
That's what I would use.
 

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