$$\eqalign{ & v = \alpha \sqrt x ,\,\,\frac{{dx}}{{dt}} = \alpha \sqrt x \Rightarrow \frac{{dx}}{{\sqrt x }} = \alpha \,dt \cr & \int\limits_0^x {\frac{{dx}}{{\sqrt x }}} = \alpha \int\limits_0^t {dt} \cr & \Rightarrow \left[ {\frac{{2\sqrt x }}{1}} \right]_0^x = \alpha \left[ t \right]_0^t \cr & \Rightarrow 2\sqrt x = \alpha t \cr & \Rightarrow x = \frac{{{\alpha ^2}}}{4}{t^2} \cr} $$

Horizontal distance of the target is $$100\,m.$$
Speed of bullet $$= 1000\,m/s$$
Time taken by bullet to cover the horizontal distance,
$$t = \frac{{100}}{{1000}} = \frac{1}{{10}}s$$
During $$\frac{1}{{10}}s,$$  the bullet will fall down vertically due to gravitational acceleration.
Therefore, height above the target, so that the bullet hit the target is
$$\eqalign{ & h = ut + \frac{1}{2}g{t^2} = \left( {0 \times \frac{1}{{10}}} \right) + \frac{1}{2} \times 10 \times {\left( {0.1} \right)^2} \cr & = 0.05\,m \cr & = 5\,cm \cr} $$

Let $$A$$ and $$B$$ be two forces. The sum of the two forces.
$${F_1} = A + B\,......\left( {\text{i}} \right)$$
The difference of the two forces,
$${F_2} = A - B\,......\left( {{\text{ii}}} \right)$$
Since, sum of the two forces is perpendicular to their differences as given, so
$$\eqalign{ & {F_1} \cdot {F_2} = 0 \cr & {\text{or}}\,\left( {A + B} \right) \cdot \left( {A - B} \right) = 0 \cr & {\text{or}}\,{A^2} - A \cdot B + B \cdot A - {B^2} = 0 \cr & {\text{or}}\,{A^2} = {B^2} \cr & {\text{or}}\,\left| A \right| = \left| B \right| \cr} $$
Thus, the forces are equal to each other in magnitude.

Angle between two vectors is given as from dot product $$A \cdot B = \left| A \right|\left| B \right|\cos \theta $$
$$\eqalign{ & \cos \theta = \frac{{A \cdot B}}{{AB}} \cr & {\text{Here,}}\,\,A = 3\hat i + 4\hat j + 5\hat k \cr & B = 3\hat i + 4\hat j - 5\hat k \cr & \therefore A = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 5 \right)}^2}} = \sqrt {50} \cr & B = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 5} \right)}^2}} = \sqrt {50} \cr & {\text{and}}\,A \cdot B = \left( {3\hat i + 4\hat j + 5\hat k} \right) \cdot \left( {3\hat i + 4\hat j - 5\hat k} \right) \cr & = 9 + 16 - 25 = 0 \cr & \therefore \cos \theta = \frac{0}{{\sqrt {50} \cdot \sqrt {50} }} = 0 \Rightarrow \theta = {90^ \circ } \cr} $$

Given, $$h = Ax - B{x^2},$$    on comparing with $$y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }},$$     we get $$A = \tan \theta = \tan {45^ \circ } = 1,$$
$$\eqalign{ & {\text{and}}\,\,B = \frac{g}{{2{u^2}{{\cos }^2}\theta }} = \frac{{10}}{{2 \times {{20}^2} \times {{\cos }^2}{{45}^ \circ }}} = \frac{1}{{40}} \cr & \therefore \frac{A}{B} = 40 \cr} $$

According to the question, at any instant $$t,$$
$$\eqalign{ & x = 4{t^2},y = 3{t^2} \cr & \therefore {v_x} = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {4{t^2}} \right) = 8t \cr & {\text{and}}\,\,{v_y} = \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {3{t^2}} \right) = 6t \cr} $$
The speed of the particle at instant $$t.$$
$$v = \sqrt {v_x^2 + v_y^2} = \sqrt {{{\left( {8t} \right)}^2} + {{\left( {6t} \right)}^2}} = 10\,t$$

Concept
Unit vector can be found by dividing a vector with its magnitude i.e. $$\hat A = \frac{A}{{\left| A \right|}}$$
Let we represent the unit vector by $$\hat n.$$ We also know that the modulus of unit vector is 1 i.e., $$\left| {\hat n} \right| = 1$$
$$\eqalign{ & \therefore \left| {\hat n} \right| = \left| {0.5\hat i + 0.8\hat j + c\hat k} \right| = 1 \cr & {\text{or}}\,\,\sqrt {{{\left( {0.5} \right)}^2} + {{\left( {0.8} \right)}^2} + {c^2}} = 1 \cr & {\text{or}}\,\,0.25 + 0.64 + {c^2} = 1 \cr & {\text{or}}\,\,0.89 + {c^2} = 1 \cr & {\text{or}}\,\,{c^2} = 1 - 0.89 = 0.11 \Rightarrow c = \sqrt {0.11} \cr} $$

Let the points $$A, B, C$$  and $$D$$ be separated by $$1\,km.$$  Then
$$\eqalign{ & {t_{AB}} = \frac{1}{{60}}hr,{t_{BC}} = \frac{1}{{40}}hr \cr & \therefore < {v_{AC}} > = \frac{{1 + 1}}{{\frac{1}{{60}} + \frac{1}{{40}}}} = 48\,km/hr \Rightarrow {t_{CD}}\frac{1}{{48}}hr. \cr & {\text{Now}}\, < {v_{AD}} > = \frac{{1 + 1 + 1}}{{\frac{1}{{60}} + \frac{1}{{40}} + \frac{1}{{48}}}} = 48\,km/hr \cr} $$

$$\eqalign{ & {\text{Given,}}\,x = 5t - 2{t^2} \cr & {\text{Velocity of the particle,}} \cr & {v_x} = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {5t - 2{t^2}} \right) = 5 - 4t \cr & {\text{Acceleration, }}{a_x} = \frac{d}{{dt}}{v_x} = - 4\;m{s^{ - 2}} \cr & {\text{Also}}\,y = 10t \cr & {\text{Velocity,}}\,{v_y} = \frac{{dy}}{{dt}} = 10 \cr & \therefore {\text{Acceleration }}{a_y} = \frac{{d{v_y}}}{{dt}} = 0 \cr & \therefore {\text{Net acceleration of the particle,}} \cr & {a_{net}} = {a_x}\widehat i + {a_y}\widehat j = \left( { - 4\;m{s^{ - 2}}} \right)\widehat i \cr & {\text{or}}\,{a_{net}} = - 4\,\widehat i\,m{s^{ - 2}} \cr} $$

The time to reach the maximum height without air resistance; $${t_1} = \frac{u}{g} = \frac{u}{{10}},$$
and with air resistance, $${t_2} = \frac{u}{{10 + 1}} = \frac{u}{{11}}$$
$$\eqalign{ & = \frac{{{t_1} - {t_2}}}{{{t_1}}} \times 100 = \left[ {\frac{{\frac{u}{{10}} - \frac{u}{{11}}}}{{\frac{u}{{10}}}}} \right] \times 100 \cr & = 9\% \left( {{\text{decreases}}} \right) \cr} $$