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A golf is launched at a speed

According to the author of the [paper][2] "The Physics of Putting" the vertical velocity is given by

$$ v_{f,v} = v_f \cdot \cos(\beta) \cdot \sin(\phi) $$

The horizontal velocity is equal to:

$$ v_{f,h} = v_f \cdot (\cos^2(\beta_f) \cos^2(\phi) + \sin^2(\beta_f))^{1/2} $$

I understand the reasoning behind the vertical velocity component. Basically we have the hypotenuse,

My (unsuccessful) attempt:

Pythagorean theorem

$$ \sin^2(\phi) + \cos^2(\phi) = \frac {v_{f,h}^2 + v_{f_v}^2} {v_f^2} $$

$$ v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) = v_{f,h}^2 + v_{f_v}^2 $$

$$ v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - v_{f_v}^2 $$

Substitute vertical velocity according to author

$$ v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2 $$

$$ v_{f,h}^2 = (v_f\cos(\beta))^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2 $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta) \cdot (\sin^2(\phi) + \cos^2(\phi)) - v_f^2\cos^2(\beta)\sin^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\sin^2(\phi) + v_f^2\cos^2(\beta)\cos^2(\phi) - v_f^2\cos^2(\beta)\sin^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\cos^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\cos^2(\phi) $$

$$ v_{f,h} = v_f(\cos^2(\beta)\cos^2(\phi))^{1/2} $$

Let me know what I am doing wrong. Thanks in advance

[2]: http://www.puttingzone.com/Science/cjp-putting.pdf

*v,f*and launch angle,*β,f*. The slope of the green is equal to*φ*. At some point the ball is located on the rim of a hole. The side view (a) and overhead view (b) looks as in the attached image.According to the author of the [paper][2] "The Physics of Putting" the vertical velocity is given by

$$ v_{f,v} = v_f \cdot \cos(\beta) \cdot \sin(\phi) $$

The horizontal velocity is equal to:

$$ v_{f,h} = v_f \cdot (\cos^2(\beta_f) \cos^2(\phi) + \sin^2(\beta_f))^{1/2} $$

I understand the reasoning behind the vertical velocity component. Basically we have the hypotenuse,

*v,fy*(*v,f * cos(β,f)*) and multiply this with*sin(φ)*. Now we have the opposite. However, why is the author not following the same logic for the horizontal velocity and multiplying the hypotenuse with*cos(φ)?*Please show me how his formulation is derived.My (unsuccessful) attempt:

Pythagorean theorem

$$ \sin^2(\phi) + \cos^2(\phi) = \frac {v_{f,h}^2 + v_{f_v}^2} {v_f^2} $$

$$ v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) = v_{f,h}^2 + v_{f_v}^2 $$

$$ v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - v_{f_v}^2 $$

Substitute vertical velocity according to author

$$ v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2 $$

*v,f*is*v,fy*$$ v_{f,h}^2 = (v_f\cos(\beta))^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2 $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta) \cdot (\sin^2(\phi) + \cos^2(\phi)) - v_f^2\cos^2(\beta)\sin^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\sin^2(\phi) + v_f^2\cos^2(\beta)\cos^2(\phi) - v_f^2\cos^2(\beta)\sin^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\cos^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\cos^2(\phi) $$

$$ v_{f,h} = v_f(\cos^2(\beta)\cos^2(\phi))^{1/2} $$

Let me know what I am doing wrong. Thanks in advance

[2]: http://www.puttingzone.com/Science/cjp-putting.pdf

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