As            R²=A² + B² + 2 AB cos \(\theta\).
In first case,
(2k + 1)² (A² + B²) =A² + B²+ 2 AB cos \(\theta\)
or 2 AB cos \(\theta\) (A² + B²) [(2k + 1)² - 1] = (A² + B²) [2k (2k + 2)] ...(1)
In second case,
(2k + 1)² (A² + B²) =A² + B² + 2 AB cos (90° -\(\theta\)) = A² + B² + 2 AB sin \(\theta\)
2 AB sin\(\theta\) = (A² + B²) [(2k + 1)²- 1] = (A² + B²) [2k (2k - 2)] ...(ii)
Dividing (ii) by (i), we have
\(tan \theta={2k-2\over2k-2}={k-1\over k+1}\)

\(\overset\rightarrow{A}\).\(\overset\rightarrow{B}\)(2\(\hat{i}\)- 3\(\hat{j}\) + \(\hat{k}\)).(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
AB cos \(\theta\) = (2)(1) + (-3)(1) + (1)(1) = 0
AB cos \(\theta\) = 0 (as A \(\neq \)0,B\(\neq \)0)
\(\therefore\) \(\theta\) = 90⁰
or, the vectors\(\overset\rightarrow{A}\)and \(\overset\rightarrow{B}\)are mutually perpendicular.

It is given that u = 40 m s^-1, vertical distance travelled in 2s is 20 m.
Using the relation y = ut sin \(\theta\) -\(\frac{1}{2}\)gt², we have
20 = 40 sin \(\theta\)\(\times\)2-\(\frac{1}{2}\)\(\times\)10\(\times\)(2)² = 80 sin\(\theta\)-20
\(\Rightarrow\) 80 sin \(\theta\)= 20 + 20 = 40 \(\Rightarrow\) sin \(\theta\)=\(\frac{40}{80}=\frac{1}{2}\)
\(\therefore\) \(\theta\) sin^-1\(\left( \frac { 1 }{ 2 } \right) \) = 30⁰
\(\therefore\) Horizontal range of projectile, R =\(\frac { { u }^{ 2 }sin2\theta }{ g } \)
\(=\frac { \left( 40 \right) ^{ 2 }\times sin\left( 2\times { 30 }^{ 0 } \right) }{ 10 } \)
\(=\frac { 1600\times sin{ 60 }^{ 0 } }{ 10 } =160\times \frac { \sqrt { 3 } }{ 2 } \)
= 138.6m.

Seconds hand of a clock completes one rotation in 60 s i.e.
T= 60 s, 8 = 2\(\pi\) rad
\(\therefore \) Angular speed, \(\omega ={\theta\over T}={2\pi \ rad \over 60 s}\)
\(={\pi\over 30}rad \ s^{-1}\)
Length of the seconds hand, R = 4 cm.
\(\therefore \) Speed of the tip of second's hand is \(v=\omega R={\pi\over 30}\times 4={2\pi\over 15}cm \ s^{-1}\)

Acceleration needed for a particle to undergo uniform circular motion is called 'centripetal acceleration'. It is directed along the radius of circular path towards its centre. Two common examples are:
(i) An electron revolving around the nucleus of an atom in a uniform circular motion experiences a centripetal acceleration on account of Coulombian electrostatic force on electron due to nucleus.
(ii) A satellite revolving around the earth in a circular orbit experiences a centripetal acceleration on account of gravitational force due to the earth.

Let the body be projected horizontally with a velocity u, from the top of a tower of height h.
Time taken to reach the ground, t = \(\sqrt{2h/g}\)
Since the initial vertical velocity is zero and there is no acceleration in the horizontal.
Thus, x = ut
x = u\(\sqrt{2h/g}\)
i.e., h = x²  h=x²\(\frac{g}{2u^2}\)
As h \(\alpha\)x², the path is a parabola.

Horizontal distance, x = 100 m
velocity,\(\upsilon \) = 500 ms^-1
Time taken to travel this distance, t = \(\frac { x }{ \upsilon } \)= \(\frac { 1 }{ 5 } \)s
Vertical distance travelled by the bullet in time \(\frac { 1 }{ 5 } \)s is  
y = u_oyt+\(\frac { 1 }{ 2 } \)gt² = 0+\(\frac { 1 }{ 2 } \)\(\times\)10\(\times\)\(\frac { 1 }{ 25 } \)=\(\frac { 1 }{ 5 } \)m
= 20cm.

The position vector, r  = \(2.0\hat { i } +{ t }^{ 2 }\hat { j } +3.0\hat { k } \)
v(t) = \(\frac { dr }{ dt } =\frac { d }{ dt } (2.0\hat { i } +{ t }^{ 2 }\hat { j } +3.0\hat { k } )\)
v(t) = \(2.0\hat { i } +2t\hat { j } \)
a(t) = \(\frac { dv(t) }{ dt } =\frac { d }{ dt } (2.0\hat { i } +2t\hat { j } )=2\hat { j } \)
At t = 2.0 s, the magnitude of the v(t) can be given as
v(t) = \(2.0\hat { i } +2\times2\hat { j } =2.0\hat { i } +4\hat { j } \)  
v = \(\sqrt { 4+16 } =\sqrt { 20 } \)  and direction is
\(\theta ={ tan }^{ -1 }\left( \frac { { { v }_{ y } } }{ { v }_{ x } } \right) ={ tan }^{ -1 }(\frac { 4 }{ 2 } )\approx { 63 }^{ o }\)

As, maximum height, H = \(\frac { { u }^{ 2 } }{ 2g } { sin }^{ 2 }\theta \)
Consider \(\triangle \)H b ethe increase in H when u changes by \(\triangle \)u, it can be obtained by differentiating the above equation, we get
\(\triangle H=\frac { 2u\triangle u{ sin }^{ 2 }\theta }{ 2g } =\frac { 2\triangle u }{ u } H\)
\(\\ \Rightarrow \frac { \triangle H }{ H } =\frac { 2\triangle u }{ u } \)
Given % increasing in H is 10% so 
\(\frac { \triangle H }{ H } =\frac { 10 }{ 100 } =0.1\Rightarrow \frac { 2\triangle u }{ u } =0.1\)
\(\\ As,\quad R=\frac { { u }^{ 2 }sin2\theta }{ g } \)
\(\\ \because \quad \triangle R=\frac { 2u\triangle u }{ u } sin2\theta \Rightarrow \frac { \triangle R }{ R } =\frac { 2\triangle u }{ u } =0.1\)
\(\\ \therefore \ increase \ in \ horizontal \ range=\frac { \triangle R }{ R } \times 100\)
= 0.1 x 100
= 10%

Given, initial velocity (u) = 40m/s
Height of the hall (H) = 25m
Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25m.
Maximum height attained by the ball
\(H=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \Rightarrow 25=\frac { { (40) }^{ 2 }{ sin }^{ 2 }\theta }{ 2\times 9.8 }\)
\(or\ { \sin }^{ 2 }\theta =\frac { 25\times 2\times 9.8 }{ 1600 } =0.3068\)
\(or\ \sin\theta =0.5534=\sin{ 33.6 }^{ 0 }\)
\(or\ \theta ={ 33.6 }^{ 0 }\)
\( \therefore \text{ Horizontal range (R)}=\frac { { u }^{ 2 }\sin2\theta }{ g }\)
\( \\ =\frac { { (40) }^{ 2 }sin2\times { 33.6 }^{ 0 } }{ 9.8 } =\frac { 1600\times sin{ 67.2 }^{ 0 } }{ 9.8 } \)
\(\\ =\frac { 1600\times 0.9219 }{ 9.8 } =150.5m\)

15⁰

\(\theta ={ 45 }^{ o }\)

Suppose the particle is projected with velocity u at an angle theta with the horizontal. Horizontal component of its velocity at all height will be \(u \cos \theta\). At the greatest height, the vertical component of velocity is zero, so the resultant velocity is
\(v_1=u \cos \theta\)
At half the greatest height during upward motion,
\(y=h / 2, a_y=-g, u_y=u \sin \theta\)
Using \(v_y^2-u_y^2=2 a_y y\)
we get, \(v_y^2-u^2 \sin ^2 \theta=2(-g) \frac{h}{2}\)
or \(v_y^2=u^2 \sin ^2 \theta-g \times \frac{u^2 \sin ^2 \theta}{2 g} \)
\(=\frac{u^2 \sin ^2 \theta}{2}\left[\because h=\frac{u^2 \sin ^2 \theta}{2 g}\right]\)
or \(v_y=\frac{u \sin \theta}{\sqrt{2}}\)
Hence, resultant velocity at half of the greatest height is
\(v_2=\left(\sqrt{v_y^2}+\left(v_y^2\right)\right)\)
\(=\left(\sqrt{u^2 \cos ^2 \theta+\left(\frac{u^2 \sin ^2 \theta}{2}\right)}\right)\)
Given, \(\frac{v_1}{v_2}=\left(\sqrt{\frac{2}{5}}\right)\)
\(\therefore \frac{v_1^2}{v_2^2}=\frac{u^2 \cos ^2 \theta}{u^2 \cos ^2 \theta+\left(\frac{u^2 \sin ^2 \theta}{2}\right)}=\frac{2}{5} \)
\(\text { or } \frac{1}{1+\frac{1}{2} \tan ^2 \theta}=\frac{2}{5}\)
or \(\frac{1}{1+\frac{1}{2} \tan ^2 \theta}=\frac{2}{5}\)
or \(2+\tan ^2 \theta=5\) or \(\tan ^2 \theta=3\)
or \(\tan \theta=(\sqrt{3})\)
\(\therefore \theta=60^{\circ} \text {. }\)