It can be observed from figure that $$P$$ and $$Q$$ shall collide if the initial component of velocity of $$P$$ on inclined plane i.e., along incline $${u_{II}} = 0$$   that is particle is projected perpendicular to incline. Time of flight on an inclined plane of inclination $$\alpha $$ is given by

$$\eqalign{ & T = \frac{{2u\sin \left( {\theta - \alpha } \right)}}{{g\cos \alpha }}, \cr & \Rightarrow 4 = \frac{{2u\sin {{90}^ \circ }}}{{10 \times \frac{1}{2}}} \Rightarrow u = 10\,m/s \cr} $$

Suppose two vectors are $$P$$ and $$Q.$$
It is given that $$\left| {P + Q} \right| = \left| {P - Q} \right|$$
Let angle between $$P$$ and $$Q$$ is $$\phi .$$
$$\eqalign{ & \therefore {P^2} + {Q^2} + 2PQ\cos \phi = {P^2} + {Q^2} - 2PQ\cos \phi \cr & \Rightarrow 4PQ\cos \phi = 0 \cr & \Rightarrow \cos \phi = 0\,\,\left[ {\because P,Q \ne 0} \right] \cr & \Rightarrow \phi = \frac{\pi }{2} = {90^ \circ } \cr} $$

At a time $$t$$ when velocity vector become mutually perpendicular
$$\eqalign{ & v\cos {45^ \circ } = 5\,{\text{horizontal component}} \cr & v = \frac{5}{{\cos {{45}^ \circ }}} = 5\sqrt 2 \,m/s \cr} $$
Vertically,
$$\eqalign{ & v\sin {45^ \circ } = gt \cr & \Rightarrow t = \frac{{v\sin {{45}^ \circ }}}{g} = \frac{{5\sqrt 2 \times \frac{1}{{\sqrt 2 }}}}{{10}} = \frac{1}{2} \cr & {\text{so,}}\,\,OA = OB = v\cos {45^ \circ } \times t = 5 \times \frac{1}{2} = 2.5 \cr & \Rightarrow AB = 2.5 \times 2 = 5\,m \cr} $$

Draw the situation as given in questions. $$OA$$  represents the path of the particle starting from origin $$O\left( {0,0} \right).$$  Draw a perpendicular from point $$A$$ to $$x$$-axis. Let path of the particle makes an angle $$\theta $$ with the $$x$$-axis, then

$$\eqalign{ & \tan \theta = {\text{slope of line }}OA \cr & \,\,\,\,\,\,\,\,\,\,\, = {\text{path of the particle }}1{\text{ making angle }}\theta \cr & \tan \theta = \frac{{3 - 0}}{{\sqrt 3 - 0}} = \sqrt 3 \cr & \,\,\,\,\,\,\theta = {60^ \circ } \cr} $$

$$\eqalign{ & \vec A.\vec B = 0 \cr & {\text{or}}\,\,\left( {5\hat i + 7\hat j - 3\hat k} \right).\left( {2\hat i + 2\hat j - a\hat k} \right) = 0 \cr & \therefore 5 \times 2 + 7 \times 2 + 3a = 0 \cr & {\text{or}}\,\,a = - 8 \cr} $$

Height attained by balls in $$2\,\sec $$  is $$ = \frac{1}{2} \times 9.8 \times 4 = 19.6\,m$$
the same distance will be covered in 2 second (for descent)
Time interval of throwing balls, remaining same. So, for two balls remaining in air, the time of ascent or descent must be greater than 2 seconds. Hence speed of balls must be greater than $$19.6\,m/\sec .$$

$$\eqalign{ & \Delta p = m\left( {v - u} \right) \cr & \left| {\Delta p} \right| = \left| {mv\left[ {\cos 45\hat i - \sin 45\hat j} \right] - mv\left[ {\cos 45\hat i + \sin 45\hat j} \right]} \right| \cr & = 2mv\sin {45^ \circ } \cr & = \sqrt 2 mv \cr} $$
Alternative
The $$\alpha $$ component of velocity will remain constant.
$$\eqalign{ & \Rightarrow \Delta {v_y} = \frac{v}{{\sqrt 2 }} - \left( {\frac{{ - v}}{{\sqrt 2 }}} \right) \cr & = \frac{{2v}}{{\sqrt 2 }} = \sqrt 2 v \cr & \Rightarrow {\text{So,}}\,\Delta {P_y} = m\sqrt 2 v = \sqrt 2 mv \cr} $$

The motions of the points are sketched in the figure. As they start moving simultaneously symmetrically, they will meet at the centroid of the triangle.

The velocity of any point towards centroid of triangle $$O$$
$$ = v\cos {30^ \circ } = \frac{{\sqrt 3 v}}{2}.$$
And its displacement $$ = \frac{{\frac{a}{2}}}{{\cos {{30}^ \circ }}} = \frac{a}{{\sqrt 3 }}.$$
Time taken to converge the points $$ = \frac{{{\text{displacement}}}}{{{\text{velocity}}}} = \frac{{2a}}{{3v}}.$$

The equation for the given $$v-x$$  graph is
$$\eqalign{ & v = - \frac{{{v_0}}}{{{x_0}}}x + {v_0}.....(i) \cr & \frac{{dv}}{{dx}} = - \frac{{{v_0}}}{{{x_0}}} \cr & \therefore v\frac{{dv}}{{dx}} = - \frac{v}{{{x_0}}} \times v \cr & = - \frac{{{v_0}}}{{{x_0}}}\left[ { - \frac{{{v_0}}}{{{x_0}}}x + {v_0}} \right]{\text{from }}(i) \cr & \therefore a = \frac{{v_0^2}}{{x_0^2}}x - \frac{{v_0^2}}{{{x_0}}}.....(ii)\,\,\left[ {\because a = v\frac{{dv}}{{dx}}} \right] \cr} $$
On comparing the equation (ii) with equation of a straight line

$$\eqalign{ & y = mx + c \cr & {\text{we get }}m = \frac{{v_0^2}}{{x_0^2}} = + ve, \cr & i.e.,\,\tan \theta = + ve,\,\,\,i.e.,\,\,\theta \,\,{\text{is acute}}{\text{.}} \cr & {\text{Also}}\,\,c = - \frac{{v_0^2}}{{x_0^2}}, \cr} $$
$$i.e.,$$   the $$y$$-intercept is negative
The above conditions are satisfied in graph (A).

$$\eqalign{ & \tan \theta = \frac{{{u^2}}}{{Rg}} \cr & \Rightarrow R = \frac{{{u^2}}}{{g\tan \theta }} = \frac{{100}}{{10 \times \sqrt 3 }} = \frac{{10}}{{\sqrt 3 }}\;m \cr} $$