\(\frac { 2u^{ 2 }\ sin^{ 2 }\ \theta }{ g } =\frac { \ u^{ 2 }\sin\ 2\theta }{ g } \)
sin²\(\theta\) = 2.2 sin \(\theta\) cos \(\theta\)      [ \(\therefore\) sin 28 = 2 sin \(\theta\) cos \(\theta\)]
sin \(\theta\) = 4 cos \(\theta\)
or,  tan \(\theta\) = 4
\(\theta\) = tan^-1(4).

T₁=  \(\frac { 2u \ sin \ \theta }{ g } \) and T₂ = \(\frac { 2u\ sin(90^{ 0 }-\theta ) }{ g } \)
Now, \(\frac { { T }_{ 1 } }{ { T }_{ 2 } } =\frac { 2u \ sin \ \theta }{ g } \times \frac { g }{ 2u\ sin(90^{ 0 }-\theta ) } \)
= sin\(\theta\) cos \(\theta\) = tan \(\theta\)

(i) The maximum height attained by the ball is 
\({ H }_{ m }=\frac { ({ \nu }_{ 0 }sin{ \theta }_{ 0 })^{ 2 } }{ 2g } \)
\(\\ =\frac { (28sin30^{ \circ })^{ 2 } }{ 2(9.8) } =\frac { 14\times 14 }{ 2\times 9.8 } =10.0\quad m\) 
(ii) The time taken by the ball to return the same level is
\(T=\left( { 2\nu }_{ 0 }sin\theta _{ 0 } \right) /g=(2\times 28\times sin30^{ \circ })/9.8\)
\(=28/9.8=2.9s\) 
(iii) The distance from the thrower to the point where the ball returns to the same level is 
\(R=\frac { \left( { \nu }_{ 0 }^{ 2 }sin2\theta _{ 0 } \right) }{ g } =\frac { 28\times 28\times sin60^{ \circ } }{ 9.8 } =69m\) 

\(\therefore \)   Ratio is 
\(\frac { A.B }{ \left| A.B \right| } =\frac { AB\quad cos\quad \theta }{ AB\quad sin\quad \theta } =cot\quad \theta \)
\(\\ =cot\quad { 60 }^{ o }=\frac { 1 }{ \sqrt { 3 } } \)
\(\\ As,\quad\theta ={ 60 }^{ o },cot{ 60 }^{ o }=\frac { 1 }{ \sqrt { 3 } } \) 

We know that \(A.B=ABcos\theta \)  when vectors are orthogonal \(\theta ={ 90 }^{ o }\)
So,    \(A.B=ABcos90^{ o }=0\)  when vectors are parallel   \(\theta ={ 0 }^{ o }\)   so,   \(A.B=ABcos{ 0 }^{ o }=AB\ (maximum)\)

As we know that
\(\left| A+B \right| =\sqrt { A^{ 2 }+{ B }^{ 2 }+2AB\quad cos\quad \theta } \)
And    \(\left| A-B \right| =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\quad \theta } \)  
But as per question, we have
\(\sqrt { A^{ 2 }+{ B }^{ 2 }+AB\quad cos\quad \theta } =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\theta } \)
Squaring both sides, we have (4 AB cos ) = 0
\(\sqrt { A^{ 2 }+{ B }^{ 2 }+AB\quad cos\quad \theta } =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\theta } \)
Hence, the two vectors A and B are perpendicular to each other.

Yes, when a man walks on the road along an oblique direction. The horizontal component of the reaction helps the man to walk on the road.

If three vectors acting on a point object at the same time are represented in magnitude and direction by tlte three sides of a triangle taken in the same order, their" resultant is zero.
The object is said to be in equilibrium

Two only, provided that they are acting in opposite directions.

We know that C² = A² + B² or 13² = 12² + 5²
Thus, the angle between A and B is 90⁰

For the vector, A_x= 2, A_y = 2
We know that angle is given by
\(\theta =tan^{ -1 }\left( \frac { A_{ y } }{ A_{ x } } \right) =tan^{ -1 }\left( \frac { 2 }{ 2 } \right)  \)
\(\theta =tan^{ -1 }(1)=45^{ 0 }\)

Consider a vector \(\overset\rightarrow{F}\)
When x, y and z components are reversed, we get
F_x\(\left( -\overrightarrow { i } \right) \)+F_y\(\left( -\overrightarrow { j } \right) \)+F_Z\(\left( -\overrightarrow { K } \right) \)=-\(\left[ { F }_{ x }\hat { i } +{ F }_{ y }\hat { j } +{ F }_{ Z }\hat { k } \right] =-\overrightarrow { F } \)
Therefore, the vector itself is reversed.

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 \(v=\omega R={\pi\over 30}\times 4={2\pi\over 15}cm \ s^{-1}\)

Subtraction of vectors is a special case of vector addition. Subtraction of \(\overset\rightarrow{B}\) from \(\overset\rightarrow{A}\) may be considered as addition of \(\left( -\overrightarrow { B } \right) \) i.e., negative of \(\overset\rightarrow{B}\) to vector \(\overset\rightarrow{A}\) .
\(\therefore\) \(\overset\rightarrow{A}\)-\(\overset\rightarrow{B}\)=\(\overset\rightarrow{A}\)+\(\left( -\overrightarrow { B } \right) \)
Therefore, we first draw \(\overset\rightarrow{A}\)and \(\overset\rightarrow{B}\)  Now draw a vector having same magnitude as of \(\overset\rightarrow{B}\)  but in oppo;ite direction. It is \(\left( -\overrightarrow { B } \right) \).sum of \(\overset\rightarrow{A}\) and \(\overset\rightarrow{B}\) gives the requisite result. In Fig.
\(\overset\rightarrow{A}\)-\(\overset\rightarrow{B}\)=\(\overset\rightarrow{A}\)+\(\left( -\overrightarrow { B } \right) \)
= \(\overset\rightarrow{KL}\)+\(\left( -\overrightarrow { LM } \right) \)
=\(\overset\rightarrow{KL}\)+\(\overset\rightarrow{LN}\)
=\(\overset\rightarrow{KL}\) .

3

(i) The magnitude of resultant R of two vectors A and B is
 Given by, 
 \(R=\sqrt { { A }^{ 2 }+B^{ 2 }+2AB\quad cos\quad \theta } \)
\(=\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+2\times 3\times 4\quad cos\quad \theta } \)
R is 7m,if \(\theta ={ 0 }^{ o }\)
(ii) The magnitude of resultant R of two vectors A and B is
 Given by,
 \(R=\sqrt { { A }^{ 2 }+B^{ 2 }+2AB\quad cos\quad \theta } \)
\(=\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+2\times 3\times 4\quad cos\quad \theta } \)
R is m,if   \(\theta ={ 180 }^{ o }\)
(iii) The magnitude of resultant R of two vectors A and B is
 Given by,
 \(R=\sqrt { { A }^{ 2 }+B^{ 2 }+2AB\quad cos\quad \theta }\)
\(=\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+2\times 3\times 4\quad cos\quad \theta } \)
R is m ,if   \(\theta ={ 90 }^{ o }\)

Let AB be the tower of height H and 0 the point of projection at a horizontal distance r from A as shown in Fig. Let u and \(\theta\) be the speed and the angle of projection of the ball.
The ball will go just grazing past the top edge of the tower if r equals the horizontal range of the projectile, i.e., if

\(r=u \ cos \theta .{2u \ sin \theta \over g}={u^2 \ sin 2\theta \over g}\)
Thus r, U and \(\theta\) must be related according to the relation
g r = u² sin 2\(\theta\)
At point A, the velocity of the ball is again u at an angle \(\theta\) to the horizontal. The horizontal distance x at which the ball strikes the ground from the foot of the tower is the horizontal distance covered with a horizontal velcoity u cos\(\theta\) in the time the ball falls vertically through a distance H starting with an initial vertically downward velocity u sin \(\theta\)and having a vertically downward acceleration g. If t is this time, we have
H=(u sin \(\theta\)) t + \({1\over2}g \ t^2\)
or      g t² + 2 u sin \(\theta\) t - 2H = 0
 or     t= \({-2 u sin \theta \pm \sqrt{4 \ u^2 sin ^2 \theta +8g H}\over 2g}\)
or     t= \({-2 u sin \theta \pm 2\sqrt{ \ u^2 sin ^2 \theta +2g H}\over 2g}\)
 t= \({1\over g}[{-u \ sin \theta \pm \sqrt{ \ u^2 sin ^2 \theta +2g H}}]\)
Thus \(x=u \ cos \theta \ t={u \ cos \theta \over g}[\sqrt{ \ u^2 sin ^2 \theta +2g H}-u\ sin \theta]\)
The angle of projection \(\theta\) for which x is minimum for a given value of u is given by
\({dx\over d \theta}=0\)
Thus, \({u \ cos \theta \over g}[{u^22sin \theta cos \theta \over \sqrt{u^2sin ^2\theta+2 \ g \ H}}-u \ cos \theta]+[\sqrt{u^2sin^2\theta+2 \ g H}-u \ sin \ \theta] ({-u \ sin \theta\over g})=0\)
or \({u^2sin2 \theta cos \theta \over \sqrt{gu^2sin ^2\theta+2 \ g \ H}}-{u \ cos^2 \theta \over g}-{ \ sin \theta\over g}\sqrt {u^2sin^2\theta+2 \ g H}+{u \ sin^2 \theta\over g}=0\)
or\({u^2sin 2\theta cos \theta \over \sqrt{u^2sin ^2\theta+2 \ g \ H}}- sin \theta \sqrt {u^2sin^2\theta+2 \ g H}-u \ cos 2 \theta=0\)
or \({u^2sin \theta cos \theta}- sin \theta \sqrt {u^2sin^2\theta+2 \ g H}-u \ cos 2 \theta\sqrt{u^2sin ^2\theta+2 \ g \ H}=0\)
The angle of projection for which x is minimum is a solution of this equation.

The vector v_b representing the velocity of the motorboat and the vector v_c representing the water current are shown in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.

We can obtain the magnitude of R using the Law of cosine
\(R=\sqrt{v_{\mathrm{b}}^{2}+v_{\mathrm{c}}^{2}+2 v_{\mathrm{b}} v_{\mathrm{c}} \cos 120^{\circ}}\)
\(=\sqrt{25^{2}+10^{2}+2 \times 25 \times 10(-1 / 2)} \cong 22 \mathrm{~km} / \mathrm{h}\)
To obtain the direction, we apply the Law of sines
\(\frac{R}{\sin \theta}=\frac{v_{c}}{\sin \phi} \text { or, } \sin \phi=\frac{v_{c}}{R} \sin \theta\)
\(=\frac{10 \times \sin 120^{\circ}}{21.8}=\frac{10 \sqrt{3}}{2 \times 21.8} \cong 0.397\)
\(\phi \cong 23.4^{\circ}\)