As ball returns to starting point so displacement is zero.

Scalar product is distributive over addition.
$$\vec i \times \vec j = - \vec j \times \vec i$$

Time of flight is independent of mass

$$t = \frac{s}{v} = \sqrt {{H^2} + {L^2}} \div \sqrt {\frac{{2H}}{g}} $$

$$\eqalign{ & \frac{{dv}}{{dt}} = - 2.5\sqrt v \,\, \Rightarrow \frac{{dv}}{{\sqrt v }} = - 2.5\,dt \cr & {\text{Integrating}},\int_{6.25}^0 {{v^{ - \frac{1}{2}}}dv = - 2.5\int_0^t {dt} } \cr & \Rightarrow \left[ {\frac{{{v^{ + \frac{1}{2}}}}}{{\left( {\frac{1}{2}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t \cr & \Rightarrow - 2{\left( {6.25} \right)^{\frac{1}{2}}} = - 2.5\,t \cr & \Rightarrow t = 2\,\sec \cr} $$

$$500\cos \theta = 250 \Rightarrow \cos \theta = \frac{1}{2}\,\,{\text{or}}\,\,\theta = {60^ \circ }.$$

Let the two balls $$P$$ and $$Q$$ meet at height $$x$$ $$m$$ from the ground after time $$ts$$ from the start. We have to find distance, $$BC = \left( {100 - x} \right)$$

For ball $$P$$
$$\eqalign{ & S = xm,u = 25\;m{s^{ - 1}},a = - g \cr & {\text{From}}\,\,S = ut + \frac{1}{2}a{t^2} \cr & x = 25t - \frac{1}{2}g{t^2}\,......\left( {\text{i}} \right) \cr} $$
For ball $$Q$$
$$\eqalign{ & S = \left( {100 - x} \right)m,u = 0,a = g \cr & \therefore 100 - x = 0 + \frac{1}{2}g{t^2}\,......\left( {{\text{ii}}} \right) \cr} $$
Adding eqns. (i) and (ii), we get
$$100 = 25t\,\,{\text{or}}\,\,t = 4\,s$$
From eqn. (i),
$$x = 25 \times 4 - \frac{1}{2} \times 9.8 \times {\left( 4 \right)^2} = 21.6\,m$$
Hence distance from the top of the tower
$$ = \left( {100 - x} \right)m = \left( {100 - 21.6\,m} \right) = 78.4\,m$$

Graphs in option (C) position-time and option (A) velocity-position are corresponding to velocity-time graph option (D) and its distance-time graph is as given below. Hence distance-time graph option (B) is incorrect.

The swimmer will cross straight if the resultant velocity of river flow and swimmer acts perpendicular to the direction of river flow. It will be so if the swimmer moves making an angle $$\alpha $$ with the upstream. i.e. goes along $$OB.$$

Since velocity in vertical direction is constant,
$$\therefore {a_y} = \frac{{d{v_y}}}{{dt}} = 0$$
The acceleration in horizontal direction,
$$\eqalign{ & {a_x} = \frac{{d{v_x}}}{{dt}} = \frac{{d\left( {a{v_0}t} \right)}}{{dt}} = a{v_0} \cr & a = \sqrt {a_x^2 + a_y^2} = \sqrt {{{\left( {a{v_0}} \right)}^2} + 0} = a{v_0} \cr} $$
The total acceleration is $$a{v_0}$$ and directed along horizontal direction.
Let $$\theta $$ is the angle that the resultant velocity makes with horizontal, then
Normal acceleration $${a_n} = a\sin \theta $$   and tangential acceleration

$$\eqalign{ & {a_t} = a\cos \theta ,\,{\text{we}}\,{\text{have}}\,x = \frac{{a{y^2}}}{{2{v_0}}} \cr & {\text{or}}\,\,y = \sqrt {\frac{{2x{v_0}}}{a}} \cr} $$
Differentiating both side of equation (iii) w.r.t. $$x,$$ we get
$$1 = \frac{a}{{2{v_0}}} \times 2y \times \frac{{dy}}{{dx}}$$

$$\eqalign{ & {\text{or}}\,\,\frac{{dy}}{{dx}} = \frac{{{v_0}}}{{ay}} = \tan \theta \cr & {\text{Now}}\,\,{a_x} = a\sin \theta = a{v_0} \times \frac{{\left( {\frac{{{v_0}}}{{ay}}} \right)}}{{\sqrt {1 + {{\left( {\frac{{{v_0}}}{{ay}}} \right)}^2}} }} \cr & = \frac{{a{v_0}}}{{\sqrt {1 + {{\left( {\frac{{ay}}{{{v_0}}}} \right)}^2}} }} \cr & {a_t} = a\cos \theta = a{v_0} \times \frac{1}{{\sqrt {1 + {{\left( {\frac{{{v_0}}}{{ay}}} \right)}^2}} }} = a{v_0} \cr & \frac{{ay}}{{\sqrt {{{\left( {ay} \right)}^2} + v_0^2} }} = \frac{{{a^2}y}}{{\sqrt {1 + {{\left( {\frac{{ay}}{{{v_0}}}} \right)}^2}} }} \cr} $$