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.
$$\therefore \frac{{{R_1}}}{{{R_2}}} = 1$$

Let $${v_r} =$$  velocity of river
$${v_{br}} =$$  velocity of boat in still water and
$$w =$$  width of river
Time taken to cross the river = $$15\,\min$$
$$ = \frac{{15}}{{60}}h = \frac{1}{4}h$$
Shortest path is taken when $${v_b}$$ is along $$AB.$$  In this case,
$$\eqalign{ & v_{br}^2 = v_r^2 + v_b^2 \cr & {\text{Now,}}\,t = \frac{w}{{{v_b}}} = \frac{w}{{\sqrt {v_{br}^2 - v_r^2} }} \cr & \therefore \frac{1}{4} = \frac{1}{{\sqrt {{5^2} - v_r^2} }} \cr & \Rightarrow {5^2} - v_r^2 = 16 \cr & \Rightarrow v_r^2 = 25 - 16 = 9 \cr & \therefore {v_r} = \sqrt 9 = 3\,km/h \cr} $$
Alternative

$$\eqalign{ & t = \frac{{15}}{{60}} = \frac{1}{4}\lambda s \cr & \Rightarrow {\text{Motion along the }}Y{\text{ - axis}}\,t = \frac{y}{{{v_b}\cos \theta }} \cr & \Rightarrow \frac{1}{4} = \frac{1}{{5\cos \theta }} \Rightarrow \cos \theta = \frac{4}{5} \cr & {\text{so,}}\,\,\sin \theta = \frac{3}{5} \cr} $$
$$ \Rightarrow $$ Motion is along the $$x$$-axis. For the boat to reach at $$B$$
$$\eqalign{ & {v_b}\sin \theta = {v_r} \cr & \Rightarrow 5 \times \frac{3}{5} = {v_r} \cr & \Rightarrow {v_r} = 3\,km/h \cr} $$
NOTE
If $${v_r} \geqslant {v_{br}},$$   the boatman can never reach at point $$B.$$

Ball $$A$$ is thrown upwards from the building. During its downward journey when it comes back to the point of throw, its speed is equal to the speed of throw. So, for the journey of both the balls from point $$A$$ to $$B.$$
We can apply $${v^2} - {u^2} = 2gh.$$
As $$u,g,h$$  are same for both the balls, $${v_A} = {v_B}$$

Given, $$H = 10\,m,u = 5\,m/s,g = 10\,m/{s^2}$$
Speed on reaching ground
$$v = \sqrt {{u^2} + 2gH} $$
Now, $$v = u + at \Rightarrow \sqrt {{u^2} + 2gH} = - u + gt$$
Time taken to reach highest point is $$t = \frac{u}{g},$$
$$\eqalign{ & \Rightarrow t = \frac{{u + \sqrt {{u^2} + 2gh} }}{g} = \frac{{nu}}{g}\,\left( {{\text{from question}}} \right) \cr & \Rightarrow 2gH = n\left( {n - 2} \right){u^2} \cr & \Rightarrow n\left( {n - 2} \right) = \frac{{2gH}}{{{u^2}}} = \frac{{2 \times 10 + 10}}{{5 \times 5}} = 8 \Rightarrow n = 4 \cr} $$

$$\eqalign{ & {\text{Here }}\vec P + b\vec R = \vec S \cr & \therefore \vec R = \frac{{\vec S - \vec P}}{b} \cr & {\text{Also}}\,\vec R = \vec Q - \vec P \cr & \therefore \frac{{\vec S - \vec P}}{b} = \vec Q - \vec P \cr & \therefore \vec S - \vec P = b\vec Q - b\vec P \cr & \therefore \vec S = b\vec Q + \left( {1 - b} \right)\vec P \cr} $$

At point $$B$$ the direction of velocity component of the projectile along $$Y$$- axis reverses.
Hence, $${\overrightarrow V _B} = 2\hat i - 3\hat j$$

Given, $$x = a{e^{ - \alpha t}} + b{e^{\beta t}}$$
Velocity $$v = \frac{{dx}}{{dt}} = - a\alpha {e^{ - \alpha t}} + b\beta {e^{\beta t}} = A + B$$
where, $$A = - a\alpha {e^{ - \alpha t}}$$
$$B = b\beta {e^{\beta t}}$$
The value of term $$A = - a\alpha {e^{ - \alpha t}}$$   decreases and of term $$B = b\beta {e^{\beta t}}$$   increases with time. As a result, velocity goes on increasing with time.

We know that, horizontal ranges for complementary angles of projection will be same. The projectiles are projected at angles $$\left( {{{45}^ \circ } - \theta } \right)$$  and $$\left( {{{45}^ \circ } + \theta } \right)$$  which are complementary to each other i.e. two angles add up to give $${90^ \circ }.$$ Hence, horizontal ranges will be equal. Thus, the required ratio is $$1:1.$$
Alternative
Horizontal range of projectile $$R = \frac{{{u^2}\sin 2\alpha }}{g}$$
$$\eqalign{ & {\text{For}}\,\,\alpha = \left( {{{45}^ \circ } - \theta } \right) \cr & {R_1} = \frac{{{u^2}\sin 2\left( {{{45}^ \circ } - \theta } \right)}}{g} \cr & = \frac{{{u^2}\sin \left( {{{90}^ \circ } - 2\theta } \right)}}{g} \cr & = \frac{{{u^2}\cos 2\theta }}{g} \cr & {\text{For}}\,\,\alpha = \left( {{{45}^ \circ } + \theta } \right), \cr & {R_2} = \frac{{{u^2}\sin 2\left( {{{45}^ \circ } + \theta } \right)}}{g} \cr & = \frac{{{u^2}\sin \left( {{{90}^ \circ } + \theta } \right)}}{g} \cr & = \frac{{{u^2}\cos 2\theta }}{g} \cr & {\text{Hence,}}\,\frac{{{R_1}}}{{{R_2}}} = \frac{1}{1} \cr & {\text{or}}\,{R_1}:{R_2} = 1:1 \cr} $$

Concept
As we know that in projectile motion only velocity of $$y$$ component change, whereas velocity of $$x$$ component remains constant.
From the figure, the $$x$$-component remains unchanged, while the $$y$$-component is reversed. Thus, the velocity at point $$B$$ is $$\left( {2\hat i - 3\hat j} \right)m{s^{ - 1}}.$$

$$\eqalign{ & {\text{Acceleration}}\,f = {f_0}\left( {1 - \frac{t}{T}} \right) \cr & {\text{or}}\,f = \frac{{dv}}{{dt}} = {f_0}\left( {1 - \frac{t}{T}} \right)\,\,\,\left[ {\because f = \frac{{dv}}{{dt}}} \right] \cr & {\text{or}}\,dv = {f_0}\left( {1 - \frac{t}{T}} \right)dt......\left( {\text{i}} \right) \cr} $$
Integrating Eq. (i) on both sides,
$$\eqalign{ & \int d v = \int {{f_0}} \left( {1 - \frac{t}{T}} \right)dt \cr & \therefore v = {f_0}t - \frac{{{f_0}}}{T} \cdot \frac{{{t^2}}}{2} + c\,......\left( {{\text{ii}}} \right) \cr} $$
where, $$c$$ is constant of integration.
Now, when $$t = 0,v = 0$$
So, from Eq. (ii), we get $$c = 0$$
$$\eqalign{ & \therefore v = {f_0}t - \frac{{{f_0}}}{T} \cdot \frac{{{t^2}}}{2}\,......\left( {{\text{iii}}} \right) \cr & {\text{As,}}\,f = {f_0}\left( {1 - \frac{t}{T}} \right) \cr & {\text{When,}}\,f = 0 \cr & 0 = {f_0}\left( {1 - \frac{t}{T}} \right) \cr & {\text{As,}}\,{f_0} \ne 0,\,{\text{so,}}\,1 - \frac{t}{T} = 0 \cr & \therefore t = T \cr} $$
Substituting, $$t = T$$  in Eq. (iii),
we get $${v_x} = {f_0}T - \frac{{{f_0}}}{T} \cdot \frac{{{T^2}}}{2} = {f_0}T - \frac{{{f_0}T}}{2} = \frac{1}{2}{f_0}T$$