Givens.
• The vertical initial velocity is 20 m/s.
,• The horizontal initial velocity is 30 ms/s.
,• The gravity is -9.8 m/s^2.
Use the following formula to find the horizontal reach of the projectile.
[tex]R=\frac{v^2_0\sin 2\theta}{g}[/tex]First, we need to find the initial speed using the components we already have.
[tex]\begin{gathered} v_0=\sqrt[]{v^2_x+v^2_y}=\sqrt[]{30^2+20^2}=\sqrt[]{900+400}=\sqrt[]{1300} \\ v_0\approx36(\frac{m}{s}) \end{gathered}[/tex]Now we need to find the direction of the initial velocity.
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{y}{x})=\tan ^{-1}(\frac{20}{30}) \\ \theta\approx33.69 \end{gathered}[/tex]Once we have the initial velocity and its direction, use the formula to find the horizontal reach.
[tex]\begin{gathered} R=\frac{(\sqrt[]{1300})^2\cdot\sin (2\cdot33.7)}{-9.8} \\ R=\frac{1300\cdot(-0.99)}{-9.8} \\ R=\frac{-1286.50}{-9.8} \\ R=131.28m \end{gathered}[/tex]Therefore, the horizontal reach is 131.28 meters, approximately.
