If $$t$$ is the time of flight, then
$$\eqalign{ & 0 = vt - \frac{1}{2}\left( {g\cos \theta } \right){t^2} \Rightarrow t = \frac{{2v}}{g}g\cos \theta . \cr & R = 0 + \frac{1}{2}\left( {g\sin \theta } \right){t^2} = \frac{1}{2}g\sin \theta {\left( {\frac{{2v}}{{g\cos \theta }}} \right)^2} \cr & = \frac{{2{v^2}}}{g}\tan \theta \sec \theta \cr} $$

The situation is depicted in figure.

Let $${v_b}$$ be the velocity of boat, $${v_r}$$ be the velocity of river and $${v_{rb}}$$ be the resultant velocity of boat.
From figure and concept of relative velocity
$$\eqalign{ & v_{rb}^2 = v_r^2 + v_b^2 \cr & \therefore {v_r} = \sqrt {v_{rb}^2 - v_b^2} = \sqrt {{{10}^2} - {8^2}} = 6\,km\,{h^{ - 1}} \cr} $$

$$\eqalign{ & {s^2} = a{t^2} + 2bt + c \cr & \therefore 2\;s\frac{{ds}}{{dt}} = 2at + 2\;b \cr & {\text{or}}\,\,\frac{{ds}}{{dt}} = \frac{{at + b}}{s},\,{\text{again}}\,{\text{differentiating}} \cr & \frac{{{d^2}\;s}}{{d{t^2}}} = \frac{{a \cdot s - \left( {at + b} \right)}}{{{s^2}}} \cdot \frac{{ds}}{{dt}} = \frac{{as - \left( {at + b} \right)\left( {\frac{{at + b}}{s}} \right)}}{{{s^2}}} \cr & \therefore \frac{{{d^2}\;s}}{{d{t^2}}} = \frac{{a{s^2} - {{\left( {at + b} \right)}^2}}}{{\;{s^3}}} \cr & \therefore a = \frac{{{d^2}\;s}}{{d{t^2}}} \propto {s^{ - 3}}. \cr} $$

Velocity of the particle is given as $$v = At + B{t^2}$$
where $$A$$ and $$B$$ are constants.
$$\eqalign{ & \Rightarrow \frac{{dx}}{{dt}} = At + B{t^2}\,\left[ {\because v = \frac{{dx}}{{dt}}} \right] \cr & \Rightarrow dx = \left( {At + B{t^2}} \right)dt \cr} $$
Integrating both sides, we get
$$\eqalign{ & \int_{{x_1}}^{{x_2}} d x = \int_1^2 {\left( {At + B{t^2}} \right)} \,dt \cr & \Rightarrow \Delta x = {x_2} - {x_1} = A\int_1^2 t dt + B\int_1^2 {{t^2}} dt \cr & = A\left[ {\frac{{{t^2}}}{2}} \right]_1^2 + B\left[ {\frac{{{t^3}}}{3}} \right]_1^2 \cr & = \frac{A}{2}\left( {{2^2} - {1^2}} \right) + \frac{B}{3}\left( {{2^3} - {1^3}} \right) \cr} $$
$$\therefore $$ Distance travelled between $$1\,s$$  and $$2\,s$$  is
$$\Delta x = \frac{A}{2} \times \left( 3 \right) + \frac{B}{3}\left( 7 \right) = \frac{{3A}}{2} + \frac{{7B}}{3}$$

For complementary angles of projection $$\left( {{{45}^ \circ } + \alpha } \right)$$   and $$\left( {{{45}^ \circ } - \alpha } \right)$$   with same initial velocity $$u,$$ range $$R$$ is same.
$${\theta _1} + {\theta _2} = \left( {{{45}^ \circ } + \alpha } \right) + \left( {{{45}^ \circ } - \alpha } \right) = {90^ \circ }$$

Angle between $${\vec A}$$ and $${\vec B}$$ is given by
$$\cos \theta = \frac{{\vec A \cdot \vec B}}{{AB}} = \frac{3}{{\sqrt {21} }}$$
The unit vector perpendicular to $${\vec A}$$ and $${\vec B}$$ is given by
$$\eqalign{ & \hat n = \frac{{\vec A \times \vec B}}{{\left| {\vec A \times \vec B} \right|}} = \frac{{\left( {3\hat i + \hat j + 2\hat k} \right) \times \left( {2\hat i - 2\hat j + 4\hat k} \right)}}{{\left| {\left( {3\hat i + \hat j + 2\hat k} \right) \times \left( {2\hat i - 2\hat j + 4\hat k} \right)} \right|}} \cr & = \frac{{\hat i - \hat j - \hat k}}{{\sqrt 3 }} \cr} $$

Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball’s velocity, the motion of ball will be only in vertical direction with respect to person for that,
$$\frac{{{v_0}}}{2} = {v_0}\cos \theta \,\,{\text{or}}\,\,\theta = {60^ \circ }$$

During $$OA,$$  acceleration $$ = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}m/{s^2}$$
During $$AB,$$  acceleration $$ = - \tan {60^ \circ } = - \sqrt 3 m/{s^2}$$
Required ratio $$ = \frac{{\frac{1}{{\sqrt 3 }}}}{{\sqrt 3 }} = \frac{1}{3}$$

Initial velocity of stone $$u = 0$$
Time to reach at ground $$t = 4\,s$$
$$\eqalign{ & {\text{Acceleration }}a = + g = 10\,m/{s^2}\,\left( {{\text{As motion of body is along the acceleration due to gravity}}} \right) \cr & \therefore {\text{Height of tower }}h = ut + \frac{1}{2}g{t^2} = \left( {0 \times 4} \right) + \frac{1}{2} \times 10 \times {4^2} = 80\,m \cr} $$

From $$v = u + at,$$   we have

$$\eqalign{ & v = 0 + \alpha {t_1} \Rightarrow {t_1} = \frac{v}{\alpha } \cr & 0 = v - \beta {t_3} \Rightarrow {t_3} = \frac{v}{\beta }. \cr} $$
$$\eqalign{ & {\text{Now,}}\,\,S = \frac{1}{2}v{t_1} + v{t_2} + \frac{1}{2}v{t_3} = \frac{{{v^2}}}{{2\alpha }} + v{t_2} + \frac{{{v^2}}}{{2\beta }} \cr & \Rightarrow {t_2} = \frac{S}{v} - \frac{v}{2}\left( {\frac{1}{\alpha } + \frac{1}{\beta }} \right). \cr & {\text{Hence,}}\,\,{t_1} + {t_2} + {t_3} = \frac{S}{v} + \frac{v}{2}\left( {\frac{1}{\alpha } + \frac{1}{\beta }} \right) \cr} $$