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Please help. I think I may be over thinking this problem because it looks very simple.

Determine the minimum initial velocity, v

x = 6m

y = 3m

X:

v

x = x

v

Y:

v

y = y

v

v

I first started solving for v

0 = v

v

This is where I get stuck at because when I try to go back and plug in

y = [(v

x = [(v

That person said to divide the first formula by the second and solve for angle first, then initial velocity. When I tried that, my answers where θ = 63.4° and v

The book says the correct answers are θ = 58.3° and v

## Homework Statement

Determine the minimum initial velocity, v

_{0}, and corresponding angle, θ_{0}needed to kick the ball just over the 3m high fence.x = 6m

y = 3m

## Homework Equations

X:

v

_{x}=v_{0x}x = x

_{0}+v_{0x}tv

_{0x}= v_{0}cosθY:

v

_{y}^{2}=v_{iy}^{2}y = y

_{0}+v_{0y}t + .5gt^{2}v

_{y}^{2}= v_{0y}^{2}+ 2g(s-s_{0})v

_{0y}= v_{0}sinθ## The Attempt at a Solution

I first started solving for v

_{0y}by setting the v_{y}equal to zero and the*y*= 3 in the third*Y*equation.0 = v

_{0y}^{2}+2(-9.81)(3)v

_{0y}= 7.67 m/sThis is where I get stuck at because when I try to go back and plug in

*v*into the third x equation, I'm still left with two variables, θ and v_{0}= v_{0y}_{0}. I don't know any other way to approach this formula. I searched online and came across a similar problem that used two different formulas I've never seen before:y = [(v

_{0})sin^{2}θ]/2gx = [(v

_{0})sin(2θ)]/gThat person said to divide the first formula by the second and solve for angle first, then initial velocity. When I tried that, my answers where θ = 63.4° and v

_{0y}= 8.58 m/s.The book says the correct answers are θ = 58.3° and v

_{0}= 9.76 m/s
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