## 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

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: $\large {v_f} = {v_0} = \sqrt {\frac{{Lg}}{{\sin \left( {2\theta } \right)}}} = \sqrt {\frac{{20 \times 32}}{{\sin \left( {2 \times 37} \right)}}} = ...{\text{meters/second}}$