Using equation to trajectory

$$\eqalign{ & - h = x\tan \left( {{0^ \circ }} \right) - \frac{{g{x^2}}}{{2\left( {2gh} \right)\left( {{{\cos }^2}{0^ \circ }} \right)}} \cr & \Rightarrow x = 2h \cr} $$

$$\eqalign{ & {\overrightarrow V _A} = 10\left( { - \hat i} \right) \cr & {\overrightarrow V _B} = 10\left( {\hat j} \right) \cr & {\overrightarrow V _{BA}} = 10\hat j + 10\hat i \cr & = 10\sqrt 2 \,km/h \cr & {\text{Distance}}\,OB = 100\cos {45^ \circ } \cr & = 50\sqrt 2 \,km \cr} $$
Time taken to each the shortest distance between $$A$$ and $$B$$ $$ = \frac{{OB}}{{\overrightarrow {{V_{BA}}} }} = \frac{{50\sqrt 2 }}{{10\sqrt 2 }} = 5\,h$$

$${V_7} = {V_5} + \frac{{3\pi }}{8} \times 2$$
To calculate $${V_5}$$ i.e. velocity at end of $$5\sec .$$
$$a = {\left( {25 - {t^2}} \right)^{\frac{1}{2}}} \Rightarrow \frac{{dv}}{{dt}} = {\left( {25 - {t^2}} \right)^{\frac{1}{2}}}$$
\[\int {dv} = \int 5 {\left( {25 - 25{{\sin }^2}\theta } \right)^{\frac{1}{2}}}\cos \theta d\theta \left\{ {\begin{array}{*{20}{l}} {t = 5\sin \theta }\\ {\frac{{dt}}{{d\theta }} = 5\cos \theta }\\ {t = 0,\theta = 0}\\ {t = 5,\theta = \frac{\pi }{2}} \end{array}} \right.\]
$$\eqalign{ & = \int_0^{\frac{\pi }{2}} 5 \times 5{\cos ^2}\theta d\theta = \frac{{25\pi }}{4} \cr & {V_7} = \frac{{25\pi }}{4} + \frac{{3\pi }}{4} = \frac{{28\pi }}{4} = 7\pi = 22\,m/s \cr} $$

Double differentiation of displacement equation gives acceleration and single differentiation gives velocity of the body.
Given, $$x = 8 + 12t - {t^3}$$
We know $$v = \frac{{dx}}{{dt}}$$  and acceleration $$a = \frac{{dv}}{{dt}}$$
$$\eqalign{ & {\text{So,}}\,v = 12 - 3{t^2}\,{\text{and}}\,a = - 6t \cr & {\text{At}}\,t = 2s \cr & v = 0\,{\text{and}}\,a = - 6 \times 2 \cr & a = - 12\,m/{s^2} \cr} $$
So, retardation of the particle $$ = 12\,m/{s^2}.$$

Let $$t$$ be the time interval of two drops. For third drop to fall
$$\eqalign{ & 5 = \frac{1}{2}g{\left( {2t} \right)^2}\,\,\left[ {{\text{As}}\,u = 0} \right] \cr & {\text{or,}}\,\frac{1}{2}g{t^2} = \frac{5}{4}\,......\left( {\text{i}} \right) \cr} $$
Let $$x$$ be the distance through which second drop falls for time $$t,$$ then
$$x = \frac{1}{2}g{t^2} = \frac{5}{4}\,m\,\,\left[ {{\text{from}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right]$$
Thus, height of second drop from ground
$$ = 5 - \frac{5}{4} = \frac{{15}}{4} = 3.75\,m$$

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 $$t = \frac{u}{g},$$
$$\eqalign{ & \Rightarrow t = \frac{{u + \sqrt {{u^2} + 2gH} }}{g} = \frac{{nu}}{g}\left( {{\text{from question}}} \right) \cr & \Rightarrow 2gH = n\left( {n - 2} \right){u^2} \cr} $$

$$\eqalign{ & {\text{Given}}\,\,\vec u = 3\hat i + 4\hat j,\,\,\,\,\vec a = 0.4\hat i + 0.3\hat j,\,\,\,\,t = 10\,s \cr & \vec v = \vec u + \vec at = 3\hat i + 4\hat j + \left( {0.4\hat i + 0.3\hat j} \right) \times 10 = 7\hat i + 7\hat j \cr & \therefore \left| {\vec v} \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,{\text{units}} \cr} $$

Unit vector along $$y$$ axis = $${\hat j},$$ so the required vector
$$\eqalign{ & = \hat j - \left[ {\left( {\hat i - 3\hat j + 2\hat k} \right) + \left( {3\hat i + 6\hat j - 7\hat k} \right)} \right] \cr & = - 4\hat i - 2\hat j + 5\hat k \cr} $$

Time taken to cross the river $$t = \frac{d}{{{v_s}\cos \theta }}$$
NOTE: For time to be minimum, $${\cos \theta }$$ = maximum
$$ \Rightarrow \theta = {0^ \circ }$$
The swimmer should swim due north.

$$\overrightarrow A = 4\hat i + 6\hat j.$$   If $${B_x}$$ and $${B_y}$$ are the components of vector along $$x$$ and $$y$$ axes, then
$$\eqalign{ & 4 + {B_x} = 10, \cr & \therefore {B_x} = 6. \cr & {\text{Also}}\,6 + {B_y} = 9, \cr & \therefore {B_y} = 3. \cr & {\text{Thus,}}\,\,B = \sqrt {B_x^2 + B_y^2} = \sqrt {{6^2} + {3^2}} = \sqrt {45} \cr} $$