Given, $$\vec A = \vec B - \vec C$$
$$\therefore \vec C = \vec B - \vec A$$
If $$\theta $$ is the angle between $${\vec A}$$ and $${\vec B},$$ then
$$\eqalign{ & {C^2} = {B^2} + {A^2} - 2AB\cos \theta \cr & \therefore \cos \theta = \frac{{{B^2} + {A^2} - {C^2}}}{{2AB}} \cr} $$

Differentiate two times and put $$x = 0.$$

As the figure drawn above shows that at points $$A$$ and $$B$$ the vertical component of velocity is $$v$$ $$\sin {60^ \circ }$$  but their directions are opposite. Hence, change in velocity.
$$\Delta v = v\sin {60^ \circ } - \left( { - v\sin {{60}^ \circ }} \right) = 2v\sin {60^ \circ } = \sqrt 3 v$$

Given, position vector of the particle at $$t = 0$$  is $$\left( {2\hat i + 3\hat j} \right)$$   and $$t = 5\,s$$  is $$\left( {13\hat i + 14\hat j} \right)$$
Average velocity vector $${v_{av}} = \frac{{{\text{Net displacement}}}}{{{\text{Time taken}}}}$$
$$\eqalign{ & = \frac{{\left( {13 - 2} \right)\hat i + \left( {14 - 3} \right)\hat j}}{5} \cr & = \frac{{11\hat i + 11\hat j}}{5} \cr & = \frac{{11}}{5}\left( {\hat i + \hat j} \right) \cr} $$

Shortest route corresponds to $${\vec v_b}$$ perpendicular to river flow

$$\eqalign{ & \therefore t = \frac{d}{{{v_b}}} = \frac{d}{{\sqrt {v_{br}^2 - v_r^2} }} \cr & or,\frac{1}{4} = \frac{1}{{\sqrt {25 - v_r^2} }} \cr & \Rightarrow {v_r} = 3\,km/h \cr} $$

Distance travelled by the body in $${n^{th}}$$ second is given by
$$\eqalign{ & {s_n} = u + \frac{a}{2}\left( {2n - 1} \right) \cr & {\text{Here,}}\,u = 0 \cr & \therefore {\text{For}}\,{4^{th}}\,s,\,{s_4} = \frac{a}{2}\left( {2 \times 4 - 1} \right) \cr & {\text{and}}\,{\text{For}}\,{3^{rd}}\,s,\,{s_3} = \frac{a}{2}\left( {2 \times 3 - 1} \right) \cr & {\text{Hence,}}\,\frac{{{s_4}}}{{{s_3}}} = \frac{{\left( {2 \times 4 - 1} \right)}}{{\left( {2 \times 3 - 1} \right)}} = \frac{7}{5} \cr} $$

From figure,
$$\tan \theta = \frac{{\sqrt 3 }}{1} = \sqrt 3 \Rightarrow \theta = {30^ \circ }$$

For downward motion: $$v=-gt$$
The velocity of the rubber ball increases in downward direction and we get a straight line between $$v$$ and $$t$$ with a negative slope.
Also applying $$y - {y_0} = ut + \frac{1}{2}a{t^2}$$
We get $$y - h = - \frac{1}{2}g{t^2}\,\, \Rightarrow y = h - \frac{1}{2}g{t^2}$$
The graph between $$y$$ and $$t$$ is a parabola with $$y = h$$  at $$t= 0.$$   As time increases $$y$$ decreases.
For upward motion:
The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same.
Here $$v=u-gt$$   where $$u$$ is the velocity just after collision. As $$t$$ increases, $$v$$ decreases. We get a straight line between $$v$$ and $$t$$ with negative slope.
Also $$y = ut - \frac{1}{2}g{t^2}$$
All these characteristics are represented by graph (B).

If two non-zero vectors are represented by the two adjacent sides of a parallelogram, then the resultant is given by the diagonal of the parallelogram passing through the point of intersection of the two vectors
$$\therefore d + e = f$$

During the complete journey acceleration $$\left( {a = g} \right)$$  remains constant close to the earth provided no air resistance