Question

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

Answer 1

here is the situation described in your problem:

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