For perpendicular vector, we have $$A \cdot B = 0$$
$$\eqalign{ & \left[ {\cos \omega t\,\hat i + \sin \omega t\,\hat j} \right]\left[ {\cos \frac{{\omega t}}{2}\hat i + \frac{{\sin \omega t}}{2}\hat j} \right] = 0 \cr & \Rightarrow \cos \omega t\cos \frac{{\omega t}}{2} + \sin \omega t\,\sin \frac{{\omega t}}{2} = 0\,\left[ {\because \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B} \right] \cr & \Rightarrow \cos \left( {\omega t - \frac{{\omega t}}{2}} \right) = 0 \Rightarrow \cos \frac{{\omega t}}{2} = 0 \cr & \Rightarrow \frac{\omega }{2} = \frac{\pi }{2} \Rightarrow t = \frac{\pi }{\omega } \cr} $$
Thus, time taken by vectors which are orthogonal to each other is $$\frac{\pi }{\omega }.$$

Range $$ = \frac{{2{u^2}}}{{g\sin \theta }} = 5\sqrt 2 \,m$$

Let $$u$$  be the velocity with which the particle is thrown and $$m$$  be the mass of the particle. Then
$$K = \frac{1}{2}m{u^2}\,.....(1)$$
At the highest point the velocity is $$u\,cos\,{60^ \circ }$$   (only the horizontal component remains, the vertical component being zero at the top-most point). Therefore kinetic energy at the highest point.
$$\eqalign{ & K' = \frac{1}{2}m{\left( {u\,cos\,{{60}^ \circ }} \right)^2} \cr & = \frac{1}{2}m{u^2}co{s^2}{60^ \circ } \cr & = \frac{K}{4}\,\left[ {{\text{From }}1} \right] \cr} $$

$$\eqalign{ & x = h\cot \theta \cr & \therefore \frac{{dx}}{{dt}} = h\frac{{d\left( {\cot \theta } \right)}}{{dt}} \cr & = h\left( { - {\text{cose}}{{\text{c}}^2}\theta } \right)\frac{{d\theta }}{{dt}} \cr & {\text{or}}\,\,v = h\,\,{\text{cose}}{{\text{c}}^2}\theta \left( {\frac{{ - d\theta }}{{dt}}} \right) \cr & = \left( {8 \times {{10}^3}} \right){\text{cose}}{{\text{c}}^2}{60^ \circ } \times 0.025 = 266.67\;m/s \cr & = 960\,km/h \cr} $$

$$\eqalign{ & \because h = \frac{1}{2}g{t^2} \cr & \therefore {h_1} = \frac{1}{2}g{\left( 5 \right)^2} = 125 \cr & {h_1} + {h_2} = \frac{1}{2}g{\left( {10} \right)^2} = 500 \Rightarrow {h_2} = 375 \cr & \therefore {h_1} + {h_2} + {h_3} = \frac{1}{2}g{\left( {15} \right)^2} = 1125 \cr & \Rightarrow {h_3} = 625 \cr & {h_2} = 3{h_1},{h_3} = 5{h_1}\,\,{\text{or}}\,\,{h_1} = \frac{{{h_2}}}{3} = \frac{{{h_3}}}{5} \cr} $$

The x-coordinate of $$P = {v_x} \times t = \omega R \times \frac{\pi }{{4\omega }} = \frac{{\pi R}}{4}$$
This horizontal distance travelled will be greater than any point on the disc between $$O$$  and $$P.$$  Therefore the landing will be in unshaded area. In the same way, the horizontal distance travelled by $$Q$$  is always less than that of any point between $$O$$  and $$R.$$  Therefore the landing will be in unshaded area.

We know that,
$$\eqalign{ & {y_m} = H = \frac{{{{\left( {u\sin \theta } \right)}^2}}}{{2g}} \cr & = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \cr & \therefore \frac{{\Delta H}}{H} = \frac{{2\Delta u}}{u}. \cr & {\text{Given}}\,\,\frac{{\Delta u}}{u} = 2\% \cr & \therefore \frac{{\Delta H}}{H} = 2 \times 2 = 4\% \cr} $$

$$s = ut + \frac{1}{2}a{t^2}$$
$$ - 65 = 12t - 5{t^2}$$   on solving we get, $$t = 5\,s$$

Let $$x$$ be the total distance between points $$P$$ and $$Q$$ and $$v$$ be the velocity of car while passing a certain middle point of $$PQ.$$  If $$a$$ is the acceleration of the car, then

For part $$PQ,$$
$$\eqalign{ & {40^2} - {30^2} = 2ax \cr & {\text{or}}\,a = \frac{{350}}{x}\,......\left( {\text{i}} \right) \cr} $$
For part $$RQ,$$
$${40^2} - {v^2} = \frac{{2ax}}{2}\,......\left( {{\text{ii}}} \right)$$
Putting value of a from Eq. (i) in Eq. (ii), we have
$$\eqalign{ & {40^2} - {v^2} = 2\left( {\frac{{350}}{x}} \right)\frac{x}{2} \cr & {\text{or}}\,\,{40^2} - {v^2} = 350\quad \cr & {\text{or}}\,\,{v^2} = 1250 \cr & \Rightarrow v = 25\sqrt 2 \,km/h \cr} $$

$$\eqalign{ & R = \frac{{{u^2}{{\sin }^2}\theta }}{g},H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \cr & {H_{\max }}\,{\text{at}}\,2\theta = {90^ \circ } \cr & {\text{so,}}\,\,{H_{\max }} = \frac{{{u^2}}}{{2g}} \cr & \frac{{{u^2}}}{{2g}} = 10 \Rightarrow {u^2} = 10g \times 2 \cr & R = \frac{{{u^2}{{\sin }^2}\theta }}{g} \Rightarrow {R_{\max }} = \frac{{{u^2}}}{g} = 20\,metre \cr} $$