Comparing the given equation with
$$y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }},$$
we get $$\tan \theta = \sqrt 3 $$

In $$\frac{T}{3}\sec ,$$  the distance travelled $$ = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} = \frac{h}{9}$$
∴ Position of the ball from the ground $$ = h - \frac{h}{9} = \frac{{8h}}{9}m$$

We know,
$$R = 4H\cot \theta \Rightarrow \cot \theta = \frac{1}{2}$$

From triangle we can say that
$$\sin \theta = \frac{2}{{\sqrt 5 }},\cos \theta = \frac{1}{{\sqrt 5 }}$$
∴ Range of projectile $$R = \frac{{2{v^2}\sin \theta \cos \theta }}{g}$$
$$ = \frac{{2{v^2}}}{g} \times \frac{2}{{\sqrt 5 }} \times \frac{1}{{\sqrt 5 }} = \frac{{4{v^2}}}{{5g}}$$

$$\eqalign{ & \frac{B}{2} = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } \,......\left( {\text{i}} \right) \cr & \therefore \tan {90^ \circ } = \frac{{B\sin \theta }}{{A + B\cos \theta }} \Rightarrow A + B\cos \theta = 0 \cr & \therefore \cos \theta = - \frac{A}{B} \cr} $$
Hence, from (i) $$\frac{{{B^2}}}{A} = {A^2} + {B^2} - 2{A^2}$$
$$\eqalign{ & \Rightarrow A = \sqrt 3 \frac{B}{2} \Rightarrow \cos \theta = - \frac{A}{B} = - \frac{{\sqrt 3 }}{2} \cr & \therefore \theta = {150^ \circ } \cr} $$

$$\eqalign{ & {v^2} - {u^2} = 2as \cr & \Rightarrow a = \frac{{{u^2}}}{{2as}} = \frac{{{{\left( {20} \right)}^2}}}{{2 \times 200}} = 1\,m/{s^2} \cr} $$

Ball $$A$$ is thrown upwards from the building. During its downward journey when it comes back to the point of throw, its speed is equal to the speed of throw. So, for the journey of both the balls from point $$A$$  to $$B.$$

We can apply $${v^2} - {u^2} = 2gh.$$
As $$u, \,g, \,h$$  are same for both the balls, $${v_A} = {v_B}$$

Let the distance between two places be $$d$$ and $${t_1}$$ is time taken by car to travel first-half length, $${t_2}$$ is time taken by car to travel second-half length. Time taken by car to travel first-half length,
$${t_1} = \frac{{\left( {\frac{d}{2}} \right)}}{{40}} = \frac{d}{{80}}$$
Time taken by car to travel second-half length,
$$\eqalign{ & {t_2} = \frac{{\left( {\frac{d}{2}} \right)}}{{60}} = \frac{d}{{120}} \cr & \therefore {\text{Total time}} = {t_1} + {t_2} \cr & = \frac{d}{{80}} + \frac{d}{{120}} \cr & = d\left( {\frac{1}{{80}} + \frac{1}{{120}}} \right) \cr & = \frac{d}{{48}} \cr & \therefore {\text{Average}}\,{\text{speed}} = \frac{d}{{{t_1} + {t_2}}} \cr & = \frac{d}{{\left( {\frac{d}{{48}}} \right)}} \cr & = 48\,km/h \cr} $$
Alternative
$$\eqalign{ & {v_{av}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}} \cr & = \frac{{2 \times 40 \times 60}}{{40 + 60}} \cr & = 48\,km/h \cr} $$

Displacement $$ = \sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {3 - 1} \right)}^2}} $$
$$\eqalign{ & = \sqrt {4 + 4} \cr & = 2\sqrt 2 \,m \cr} $$

Given, initial velocity $$\left( u \right) = 3\hat i + 4\hat j$$
Final velocity $$\left( v \right) = ?$$
Acceleration $$\left( a \right) = \left( {0.4\hat i + 0.3\hat j} \right)$$
Time $$\left( t \right) = 10\,s$$
From first equation of motion, $$v = u + at$$
$$\eqalign{ & v = 3\hat i + 4\hat j + 10\left( {0.4\hat i + 0.3\hat j} \right) \cr & v = 7\hat i + 7\hat j \cr & \Rightarrow \left| v \right| = 7\sqrt 2 \cr} $$

Instantaneous velocity is the slope of displacement-time graph. At point $$E,$$ the slope is negative so instantaneous velocity of the particle is negative. At points $$C$$ and $$F,$$ the slope is positive and at $$D,$$ the slope is zero.