# Find minimal speed for ball to clear wall

1. Sep 20, 2010

### leonne

1. The problem statement, all variables and given/known data
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

2. Relevant equations
rvac= ((2Vxo)(Vyo))/g

3. 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

2. Sep 20, 2010

### kuruman

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.

3. Sep 20, 2010

### leonne

o ok thanks for the info

4. Sep 21, 2010

### leonne

so would you use this formula and solve for initial velocity? y= (x(Voy)/Voc) - 1/2 g (x^2 /V^2ox)

5. Sep 21, 2010

### kuruman

That's what I would use.