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.
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