Find minimal speed for ball to clear wall

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To determine the minimum speed for a golf ball to clear a wall of height h at a distance d, the initial velocity must be calculated using the components of the velocity vector, Vxo and Vyo. The formula provided in the discussion, rvac = (2Vxo)(Vyo)/g, is appropriate for this scenario, as it accounts for the vertical height the ball must achieve. The standard range formula, R = v^2/g sin(2@), is not applicable here since it assumes the projectile lands at the same height from which it was launched. The discussion emphasizes the need to solve for the initial velocity using the correct projectile motion equations. Understanding the relationship between the components of velocity and the wall's height is crucial for accurate calculations.
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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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