Two particles are projected with same initial velocities at an angle $${30^ \circ }$$ and $${60^ \circ }$$ with the horizontal. Then,

Solution :
(a) Maximum height in case of projectile is given by
$$\eqalign{ & H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \Rightarrow \frac{{{H_1}}}{{{H_2}}} = \frac{{{{\sin }^2}{\theta _1}}}{{{{\sin }^2}{\theta _2}}} = \frac{{{{\sin }^2}{{30}^ \circ }}}{{{{\sin }^2}{{60}^ \circ }}} \cr & = \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} = \frac{1}{3} \cr} $$
(b) Range $$R = \frac{{{u^2}\sin 2\theta }}{g}$$
$$\eqalign{ & \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{\sin \left( {2 \times {{30}^ \circ }} \right)}}{{\sin \left( {2 \times {{60}^ \circ }} \right)}} \cr & = \frac{{\sin {{60}^ \circ }}}{{\sin {{120}^ \circ }}} = \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2}}} = 1 \cr & \Rightarrow {R_1} = {R_2} \cr} $$
(c) Time of flight $$T = \frac{{2u\sin \theta }}{g}$$
$$\eqalign{ & \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{{\sin {\theta _1}}}{{\sin {\theta _2}}} \cr & = \frac{{\sin {{30}^ \circ }}}{{\sin {{60}^ \circ }}} = \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{1}{{\sqrt 3 }} \cr} $$
Hence, their horizontal ranges will be equal.
Alternative
In this problem, it is given that two particles are projected at angles $${{{30}^ \circ }}$$ and $${{{60}^ \circ }}$$ which are complementary angles. We know that horizontal range will be same for complementary angles. Hence, their ranges will be equal.