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## anonymous one year ago you throw a ball at 37 degrees above the horizontal and you want to hit a target on the ground, 20 ft from where you are standing. how fast should you throw the ball? what will be its final velocity

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1. Michele_Laino

here is the situation described in your problem: |dw:1435648400570:dw| we can apply this formula: $\Large L = \frac{{v_0^2}}{g}\sin \left( {2\theta } \right)$ where L=20 feet and g= 32 feet/sec^2 (earth gravity) The final speed v_f is equal to the initial speed v_0, so we can write: $\Large {v_f} = {v_0}$ Now from the first equation we get: $\large {v_f} = {v_0} = \sqrt {\frac{{Lg}}{{\sin \left( {2\theta } \right)}}} = \sqrt {\frac{{20 \times 32}}{{\sin \left( {2 \times 37} \right)}}} = ...{\text{meters/second}}$

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