• anonymous
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
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
  • 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: \[\large {v_f} = {v_0} = \sqrt {\frac{{Lg}}{{\sin \left( {2\theta } \right)}}} = \sqrt {\frac{{20 \times 32}}{{\sin \left( {2 \times 37} \right)}}} = ...{\text{meters/second}}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.