question_answer

What will be the effect on maximum height of a projectile when its angle of projection is changed from \[\mathbf{30}{}^\circ \] to\[\mathbf{60}{}^\circ \], keeping the same initial velocity of projection?

Answer:

                Maximum height, \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]                 \[\therefore \]  \[\frac{{{H}_{2}}}{{{H}_{1}}}=\frac{{{\sin }^{2}}{{60}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}}=\frac{{{(\sqrt{3}/2)}^{2}}}{{{(1/2)}^{2}}}=3\] Thus the maximum height becomes three times the original maximum height.