Two bodies of same mass are projected with the same velocity at an angle $${30^ \circ }$$ and $${60^ \circ }$$ respectively. The ratio of their horizontal ranges will be

Solution :
When an object is projected with velocity $$u$$ making an angle $$\theta $$ with the horizontal direction,then horizontal range will be
$${R_1} = \frac{{{u^2}\sin 2\theta }}{g}\,......\left( {\text{i}} \right)$$
when an object is projected with velocity $$u$$ making an angle $$\left( {{{90}^ \circ } - \theta } \right)$$   with the horizontal direction, then horizontal range will be
$$\eqalign{ & {R_2} = \frac{{{u^2}\sin 2\left( {{{90}^ \circ } - \theta } \right)}}{g} = \frac{{{u^2}}}{g}\sin \left( {{{180}^ \circ } - 2\theta } \right) \cr & = \frac{{{u^2}}}{g}\sin 2\theta \,......\left( {{\text{ii}}} \right) \cr} $$
From Eqs. (i) and (ii), we note that $${R_1} = {R_2}$$
Here, the projection angle is $${30^ \circ }$$ and $${60^ \circ } = \left( {{{90}^ \circ } - {{30}^ \circ }} \right),$$    so horizontal range is same for both angles.
$$\therefore \frac{{{R_1}}}{{{R_2}}} = 1$$