The angle for which the ranges are same is complementary.
Let one angle be $$\theta ,$$ then other is $${90^ \circ } - \theta $$
$$\eqalign{ & {T_1} = \frac{{2u\,\sin \theta }}{g},\,\,\,{T_2} = \frac{{2u\,\cos \theta }}{g} \cr & {T_1}{T_2} = \frac{{4{u^2}\,\sin \theta \cos \theta }}{g} = 2R\,\,\,\left( {\because R = \frac{{{u^2}{{\sin }^2}\theta }}{g}} \right) \cr} $$
Hence it is proportional to $$R.$$

Given, $$r = 7\hat i + 3\hat j + \hat k,F = - 3\hat i + \hat j + 5\hat k$$
$$\therefore \tau = r \times F = \left| r \right|\left| F \right|\sin \theta $$
where, $$\theta $$ is the angle between $$r$$ and $$F$$
$$ = \left( {7\hat i + 3\hat j + \hat k} \right) \times \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
\[ = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 7&3&1\\ { - 3}&1&5 \end{array}} \right| = \hat i\left( {15 - 1} \right) - \hat j\left( {35 + 3} \right) + \hat k\left( {7 + 9} \right)\]
$$ = 14\hat i - 38\hat j + 16\hat k$$
Alternative
$$\eqalign{ & \therefore \tau = r \times F \cr & = \left( {7\hat i + 3\hat j + \hat k} \right) \times \left( { - 3\hat i + \hat j + 5\hat k} \right) \cr & = - 21\left( {\hat i \times \hat i} \right) + 7\left( {\hat i \times \hat j} \right) + 35\left( {\hat i \times \hat k} \right) - 9\left( {\hat j \times \hat i} \right) + 3\left( {\hat j \times \hat j} \right) + 15\left( {\hat j \times \hat k} \right) - 3\left( {\hat k \times \hat i} \right) + \left( {\hat k \times \hat j} \right) + 5\left( {\hat k \times \hat k} \right) \cr & = 0 + 7\hat k - 35\hat j + 9\hat k + 0 + 15\hat i - 3\hat j - 1 + 0 \cr & = 14\hat i - 38\hat j + 16\hat k \cr} $$

The bullet performs a horizontal journey of $$100\,cm$$  with constant velocity of $$1500\,m/s.$$   The bullet also performs a vertical journey of $$h$$ with zero initial velocity and downward acceleration $$g.$$
$$\therefore $$ For horizontal journey, time $$\left( t \right) = \frac{{{\text{Distance}}}}{{{\text{Velocity}}}}$$
$$\therefore t = \frac{{100}}{{1500}} = \frac{1}{{15}}\sec \,......\left( 1 \right)$$
The bullet performs vertical journey for this time.
For vertical journey, $$h = ut + \frac{1}{2}g{t^2}$$
$$\eqalign{ & h = 0 + \frac{1}{2} \times 10 \times {\left( {\frac{1}{{15}}} \right)^2} \cr & {\text{or,}}\,\,h = \frac{{20}}{9}cm = 2.2\,cm \cr} $$

$$\eqalign{ & \frac{{{\text{Time of}}\,{\text{ascent}}}}{{{\text{Time of descent}}}} = \frac{{\left( {\frac{u}{{g + a}}} \right)}}{{\frac{u}{{\sqrt {\left( {g + a} \right)\left( {g - a} \right)} }}}} \cr & = \frac{{\sqrt {\left( {g + a} \right)\left( {g - a} \right)} }}{{\left( {g + a} \right)}} \cr & = \sqrt {\frac{{g - a}}{{g + a}}} \cr & = \sqrt {\frac{{10 - 5}}{{10 + 5}}} \cr & = \sqrt {\frac{5}{{15}}} \cr & = \sqrt {\frac{1}{3}} \cr} $$

$$\eqalign{ & {S_n} = u + \frac{a}{2}\left( {2n - 1} \right) \cr & {\text{or,}}\,\,s = \frac{a}{2}\left( {2 \times 2 - 1} \right) \Rightarrow a = \frac{2}{3}s\,m/{s^2} \cr} $$

Let $$u$$ be the initial velocity and $$H$$ be the maximum height attained.
$$\eqalign{ & {\text{At height }}h = \frac{H}{2},\,{\text{we have }}v = {v_1}\, = 10\,m/s \cr & {\text{From third equation of motion, }}v_1^2 = {u^2} - 2gh\left( {{\text{Negative sign indicates that velocity and acceleration are in opposite direction}}} \right) \cr & {\text{or}}\,{\left( {10} \right)^2} = {u^2} - 2g\frac{H}{2}\,......\left( {\text{i}} \right) \cr & {\text{At}}\,{\text{height}}\,H,\,{v_2} = 0 \cr & v_2^2 = {u^2} - 2gH\,\,{\text{or}}\,\,0 = {u^2} - 2gH\,......\left( {{\text{ii}}} \right) \cr & {\text{Subtract Eq}}{\text{. }}\left( {{\text{ii}}} \right){\text{ from Eq}}{\text{. }}\left( {\text{i}} \right),{\text{ we get}} \cr & {\left( {10} \right)^2} = 2g\frac{H}{2}\,\,{\text{or}}\,\,H = \frac{{{{\left( {10} \right)}^2}}}{g} \cr & {\text{or}}\,H = \frac{{{{\left( {10} \right)}^2}}}{{10}} = 10\,m \cr} $$
Alternative
Maximum height attained by the stone
$$\eqalign{ & H = \frac{{{u^2}}}{{2g}} \cr & {\text{When,}}\,H = \frac{H}{2},u = 10\,m/s \cr & \frac{H}{2} = \frac{{{{\left( {10} \right)}^2}}}{{2g}}\,\,{\text{or}}\,\,H = \frac{{100}}{{10}} = 10\,m \cr} $$

Clearly distance moved by 1^st ball in $$18\,s$$  = distance moved by 2^nd ball in $$12\,s.$$
Now, distance moved in $$18\,s$$  by 1^st ball $$ = \frac{1}{2} \times 10 \times {18^2} = 90 \times 18 = 1620\,\,m$$
Distance moved in $$12\,s$$  by 2^nd ball $$ = ut + \frac{1}{2}g{t^2}$$
$$\eqalign{ & \therefore 1620 = 12v + 5 \times 144 \cr & \Rightarrow v = 135 - 60 = 75\,m{s^{ - 1}} \cr} $$

Two bodies will collide at the highest point if both cover the same vertical height in the same time.

$$\eqalign{ & \Rightarrow \frac{{{V_2}}}{{{V_1}}} = \sin {30^ \circ } = \frac{1}{2} \cr & \therefore {V_2} = \frac{1}{2}{V_1} \cr} $$