$$\eqalign{ & {\overrightarrow v _{SB}} = v\hat j = {\overrightarrow v _s} + 3\hat i \cr & {\overrightarrow v _S} = v\hat j - 3\hat i\,{\text{and}}\,v = \frac{{100}}{{50}} = 2\,m/s \cr & \therefore \left| {{{\overrightarrow v }_S}} \right| = \sqrt {{v^2} + {{\left( 3 \right)}^2}} = \sqrt {{2^2} + 9} = \sqrt {13} \,m/s \cr & {\text{Drift}} = 50 \times 3 = 150\,m \cr} $$

For free fall from a height, $$u = 0$$  (initial velocity).
From second equation of motion
$$\eqalign{ & h = ut + \frac{1}{2}g{t^2} \cr & {\text{or}}\,\,h = 0 + \frac{1}{2}g{t^2} \cr & \therefore \frac{{{h_1}}}{{{h_2}}} = {\left( {\frac{{{t_1}}}{{{t_2}}}} \right)^2} \cr & {\text{Given,}}\,{h_1} = 16\;m,{h_2} = 25\;m \cr & \therefore \frac{{{t_1}}}{{{t_2}}} = \sqrt {\frac{{{h_1}}}{{{h_2}}}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5} \cr} $$
NOTE
Time taken by the object in falling does not depend on mass of object.

Ranges for complementary angles are same
∴ Required Required $$ = \frac{\pi }{2} - \frac{{5\pi }}{{36}} = \frac{{13\pi }}{{36}}$$

Let $$u$$ be the initial velocity that have to find and $$a$$ be the uniform acceleration of the particle.
For $$t = 3\,s,$$   distance travelled $$S = 12\,m$$   and for $$t = 3 + 3 = 6\,s$$    distance travelled $$S' = 12 + 30 = 42\,m$$
From, $$S = ut + \frac{1}{2}a{t^2}$$
$$\eqalign{ & 12 = u \times 3 + \frac{1}{2} \times a \times {3^2} \cr & {\text{or}}\,\,24 = 6u + 9a\,......\left( {\text{i}} \right) \cr} $$
Similarly, $$42 = u \times 6 + \frac{1}{2} \times a \times {6^2}$$
$${\text{or}}\,\,42 = 6u + 18a\,......\left( {{\text{ii}}} \right)$$
On solving, we get $$u = 1\,m{s^{ - 1}}$$

$$\eqalign{ & v_B^2 = v_T^2 + 2\left( {10} \right)\left( 3 \right) \cr & \Rightarrow v_B^2 = v_T^2 + 60\,......\left( {\text{i}} \right) \cr & {\text{Also,}}\,{v_B} = {v_T} + \left( {10} \right)\left( {0.5} \right)\,......\left( {{\text{ii}}} \right) \cr} $$
Solve eqs. (i) and (ii)
we get
$${v_T} - {v_B} = 4.9\,m{s^{ - 1}}$$

$$\eqalign{ & R = \frac{{{u^2}{{\sin }^2}\theta }}{g},\,H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}};\,{H_{\max }}{\text{ at }}2\theta = 90 \cr & {H_{\max }} = \frac{{{u^2}}}{{2g}};\,\, = 10\,\,\,\, \Rightarrow {u^2} = 10\,g \times 2 \cr & R = \frac{{{u^2}\sin 2\theta }}{{\left( g \right)}}\,\,\,\, \Rightarrow {R_{\max }} = \frac{{{u^2}}}{g} \cr & {R_{\max }} = \frac{{10 \times g \times 2}}{g} = 20\,{\text{meter}} \cr} $$

(a) Maximum height in case of projectile is given by
$$\eqalign{ & H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \Rightarrow \frac{{{H_1}}}{{{H_2}}} = \frac{{{{\sin }^2}{\theta _1}}}{{{{\sin }^2}{\theta _2}}} = \frac{{{{\sin }^2}{{30}^ \circ }}}{{{{\sin }^2}{{60}^ \circ }}} \cr & = \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} = \frac{1}{3} \cr} $$
(b) Range $$R = \frac{{{u^2}\sin 2\theta }}{g}$$
$$\eqalign{ & \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{\sin \left( {2 \times {{30}^ \circ }} \right)}}{{\sin \left( {2 \times {{60}^ \circ }} \right)}} \cr & = \frac{{\sin {{60}^ \circ }}}{{\sin {{120}^ \circ }}} = \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2}}} = 1 \cr & \Rightarrow {R_1} = {R_2} \cr} $$
(c) Time of flight $$T = \frac{{2u\sin \theta }}{g}$$
$$\eqalign{ & \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{{\sin {\theta _1}}}{{\sin {\theta _2}}} \cr & = \frac{{\sin {{30}^ \circ }}}{{\sin {{60}^ \circ }}} = \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{1}{{\sqrt 3 }} \cr} $$
Hence, their horizontal ranges will be equal.
Alternative
In this problem, it is given that two particles are projected at angles $${{{30}^ \circ }}$$ and $${{{60}^ \circ }}$$ which are complementary angles. We know that horizontal range will be same for complementary angles. Hence, their ranges will be equal.

KEY CONCEPT
Before hitting the ground, the velocity $$v$$ is given by $$v{^2} = 2gd$$   (quadratic equation and hence parabolic path)
Downwards direction means negative velocity. After collision, the direction become positive and velocity decreases.
Further,
$$\eqalign{ & v{'^2} = 2g \times \left( {\frac{d}{2}} \right) = gd; \cr & \therefore \left( {\frac{v}{{v'}}} \right) = \sqrt 2 \,\,\,or,v = v'\sqrt 2 \cr & \Rightarrow v' = \frac{v}{{\sqrt 2 }} \cr} $$
As the direction is reversed and speed is decreased graph (A) represents these conditions correctly.

Relative velocity
$$ = \left( {3\hat i + 4\hat j} \right) - \left( { - 3\hat i - 4\hat j} \right) = 6\hat i + 8\hat j.$$

At $$t=0,$$  the relative velocity will be zero.
At $$t = \frac{T}{4},$$  the relative velocity will be maximum in magnitude.
At $$t = \frac{T}{2},$$  the relative velocity will be zero.
At $$t = \frac{3T}{4},$$  the relative velocity will be maximum in magnitude
At $$t=T,$$  the relative velocity again becomes zero.