Let OAB be the inclined plane making an angle \(\alpha\) to the horizontal and let the projectile hit it at a point A where OA = r (Fig.).
At the instant the projectile hits the inclined plane, its horizontal and vertical displacements are ON (= r cos \(\alpha\)) and NA (= r sin \(\alpha\)) respectively.
Now the time taken to cover a horizontal distance r cos \(\alpha\) is clearly r cos \(\alpha\)/ v cos \(\theta\). The vertical distance moved in this time being r sin \(\alpha\), we have

\(r \ sin \alpha=(v \ sin \theta)({r \ cos \alpha \over v \ cos \theta})-{1\over2}g({r \ cos \alpha\over v cos \theta})^2\)
or \(r(sin \alpha - tan \theta cos \alpha)=-{1\over2}g{\cos^2\alpha\over v^2 \cos^2 \theta }r^2\)
\(\therefore r={2v^2cos^2\theta\over g \ cos^2 \alpha}[tan \theta cos \alpha-sin \alpha]\)
\(={2v^2cos^2\theta\over g \ cos^2 \alpha}[{sin \theta cos \alpha-cos \theta sin \alpha\over cos \theta}]\)
\(={2v^2cos^2\theta\over g \ cos^2 \alpha} sin (\theta - \alpha)\)
Thus the range of the projectile on the inclined plane is
\(={2v^2cos^2\theta\over g \ cos^2 \alpha} sin (\theta - \alpha)\)
The time at which the projectile hits the inclined plane being (r cos \(\alpha\) / v cos \(\theta\)), we have
\(t={r \ cos \alpha \over v \ cos \theta}={2v^2cos\theta\over g \ cos^2\alpha} sin (\theta - \alpha){cos \alpha\over v cos \theta}\)
or \(t={2 \ v \ sin (\theta - \alpha)\over g \ cos \alpha}\)
A confirmation of these results is obtained by putting \(\alpha\) = 0 for which case we get
\(r={2 \ v^2 cos \theta \ sin \theta \over g}={v^2sin 2\theta\over g}and \ t={2 \ v \ sin \theta \over g}\)
the usual expressions for the range and time of flight of a projectile.

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.

Let two vectors \(\vec{A}\) and \(\vec{B}\) be acting simultaneously at a particle, inclined at an angle \(\theta\) from one another. The magnitude of resultant vector \(\vec{R}\) is given by 
\(R=\sqrt{A^2+B^2+2AB \ cos \theta}\)
If the resultant vector subtends an angle B from the direction of \(\vec{A}\), then 
\(tan \beta ={B \ sin \theta\over A+B \ cos \theta}\) 
There are three special cases which are as follows
(i) If \(\theta\) = 0° i.e., vectors \(\vec{A}\)and \(\vec{B}\) are acting in same direction, then cos \(\theta\) = cos 0° = 1 and sin \(\theta\) = sin 0° = 0. Hence,
\(R=\sqrt{A^2+B^2+2AB (1)}=A+B\)
and \(tan \beta ={B \times (0)\over A+B (1)}=0 or \beta=tan ^{-1}(0)=0^0\)
Thus the magnitude of resultant vector is equal to the sum of the magnitudes of \(\vec{A}\) and \(\vec{B}\) and the resultant vector acts in the direction of \(\vec{A}\) or \(\vec{B}\).
(ii) If \(\theta\) = 180°, i.e., vectors \(\vec{A}\) and \(\vec{B}\) are acting in mutually opposite direction, then
cos \(\theta\) = cos 180° = - 1 and sin \(\theta\) = sin 180° = 0.
\(\therefore R=\sqrt{A^2+B^2+2AB (-1)}= \sqrt{A^2+B^2+2AB}=(A-B)\)
and \(tan \beta ={B \times (0)\over A+B \times(1)}=0 or \beta=tan ^{-1}(0)=0^0 \ or \ 180^o.\)
Hence the magnitude of resultant vector is equal to the difference of the magnitudes of \(\vec{A}\) and \(\vec{B}\) and the resultant vector acts in the direction of bigger of two vectors \(\vec{A}\) or \(\vec{B}\).
(iii) If\(\theta\) = 90° i.e., vectors \(\vec{A}\) and \(\vec{B}\) are in mutually perpendicular directions, then cos \(\theta\) = cos 90° = 0 and sin \(\theta\)= sin 90° = 1
\(\therefore R=\sqrt{A^2+B^2+2AB (0)}= \sqrt{A^2+B^2}\)
and \(tan \beta ={B \times (1)\over A+B .(0)}={B\over A}\)
Thus, the magnitude of resultant vector is R =\(\sqrt{A^2+B^2}\) and it subtends an angle \(\beta\) from \(\vec{A}\) such that tan\(\beta={B\over A}\) .

(i) The path followed by the motorist will be a closed hexagonal path.
Suppose the motorist starts his journey from the point O. He takes the turn at the point C.
Displacement = \(\overset\rightarrow{OC}\)
Here   OC = \(\sqrt { \left( OB \right) ^{ 2 }+\left( BC \right) ^{ 2 } } =\sqrt { \left( OF+FB \right) ^{ 2 }+\left( BC \right) ^{ 2 } } \)
= \(\sqrt { \left( 500cos\quad 30^{ 0 }+500cos30^{ 0 } \right) ^{ 2 }+(500)^{ 2 } } \)
=\(\sqrt { \left( 2\times 500\times \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }+(500)^{ 2 } } \)
= 500\(\sqrt{4}\) = 1000 m = 1 km
Total path length= 500 m + 500 m + 500 m = 1500 m = 1.5 km
\(\frac{Magnitude \ of\ displacement}{Total\ path \ length}\)=\(\frac{1}{1.5}\)=\(\frac{2}{3}\)=0.
(ii) The motorist will take the sixth turn at O.
Displacement is zero. So, displacement vector is a null vector.
Path length is 3000 m, i.e., 3 km.
Ratio of magnitude of displacement and path length is zero.
(iii) The motorist will take the 8^th turn at B.
Magnitude of displacement = 2 x 500 cos 30°= 500\(\sqrt{3m}\)=\(\frac { \sqrt { 3 } }{ 2 } \) km = 0.866 km
Path length = 8 x 500 m = 4 km 
Ratio of magnitude of displacement and path length is \(\frac { \sqrt { 3 } /2 }{ 4 } \)i.e.,\(\frac { \sqrt { 3 } }{ 2 } \) = 0.22

m =10 kg, d = 0.4m, r = 0.2m
Revolutions per min, \(\nu =1000/min=\frac { 1000 }{ 60 } s,Linear\)
velocity, v=?,centripetal acceleration, a=?
\(\\ \omega =2\pi \nu =2\pi \times \frac { 1000 }{ 60 } =\frac { 100\pi }{ 3 } rad/s\)
\(\\ v=r\omega =0.2\times \frac { 100\pi }{ 3 } =\frac { 20\pi }{ 3 } m/s\)
\(\\ a=r\omega ^{ 2 }=0.2\times \left( \frac { 100\pi }{ 3 } \right) ^{ 2 }=\frac { 2000\pi ^{ 2 } }{ 9 } m/s^{ 2 }\)

(i) \(T=\frac { { 2u }_{ 0 }\sin { \beta } }{ g\cos { \alpha } } \)
(ii) \(R=\frac { { 2u }_{ 0 }\sin { \beta } \cos { \left( \alpha +\beta \right) } }{ g\cos { ^{ 2 }\alpha } } \)
(iii) \(\beta =\frac { \pi }{ 4 } -\frac { \alpha }{ 2 } \)

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
\(\frac{R}{\sin \theta}=\frac{v_{c}}{\sin \phi} \text { or, } \sin \phi=\frac{v_{c}}{R} \sin \theta\)
\(=\frac{10 \times \sin 120^{\circ}}{21.8}=\frac{10 \sqrt{3}}{2 \times 21.8} \cong 0.397\)
\(\phi \cong 23.4^{\circ}\)

Given, A = 20 N, B = 20 N, R = 30 N, \(\theta \) = ?
 \(R=\sqrt { { A }^{ 2 }+{ B }^{ 2 }+2AB\cos { \theta } }\)
\( \\ 30=\sqrt { { 2 }0^{ 2 }+{ 2 }0^{ 2 }+2\times 20\times 20\cos { \theta } } \)
\(\\ \cos { \theta } =\frac { { 30 }^{ 2 }-{ 2 }0^{ 2 }-{ 2 }0^{ 2 } }{ 2\times { 2 }0^{ 2 } } =0.125\)
\( \Rightarrow \quad \theta ={ 82 }^{ 0 }{ 49 }^{ ' }\) 

(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 
\(R=\frac { \left( { \nu }_{ 0 }^{ 2 }sin2\theta _{ 0 } \right) }{ g } =\frac { 28\times 28\times sin60^{ \circ } }{ 9.8 } =69m\) 

\(\theta ={ 30 }^{ \circ },u=20m/s,g=10m/s^{ 2 }\)
\(\\ H=\frac { u^{ 2 }\sin { ^{ 2 }\theta } }{ 2g } =\frac { (20)^{ 2 }\times \sin { ^{ 2 }30^{ \circ } } }{ 2\times 10 } =5m\)

Yes, Multiplying any two scalars is meaningful. Density p and volume V both the scalar quantities.When density is multiplied by volume,then we get \(p\times VB=m\)   , mass of the body,which is scalar quantity