question_answer

What are the two angles of projection of a projectile projected with velocity 30 m/s, so that the horizontal range is 45 m. Take,\[~\text{g}=\text{1}0\text{ m}/{{\text{s}}^{\text{2}}}\].

Answer:

                Horizontal range = 45 = \[\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{30}^{2}}\sin 2\theta }{10}\]                 \[\sin 2\theta =\frac{45\times 10}{30\times 30}=\frac{1}{2}=\sin {{30}^{o}}\]or\[\sin {{150}^{o}}\] \[\text{2}\theta =\text{3}0{}^\circ \]or\[\text{15}0{}^\circ ~\]or\[\theta =\text{15}{}^\circ \text{ or 75}{}^\circ \]