$$\eqalign{ & x = \alpha {t^3}\,{\text{and}}\,y = \beta {t^3} \cr & {v_x} = \frac{{dx}}{{dt}} = 3\alpha {t^2}{\text{ and }}{v_y} = \frac{{dy}}{{dt}} = 3\beta {t^2} \cr & \therefore v = \sqrt {v_x^2 + v_y^2} \cr & = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}} \cr & = 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} \cr} $$

If the ball hits the $${n^{th}}$$ step, then the horizontal distance traversed $$= nh.$$  Here, the velocity along the horizontal direction $$= u.$$  Initial velocity along the vertical direction $$= 0.$$

$$\eqalign{ & {\text{So}}\,nb = ut\,......\left( {\text{i}} \right) \cr & nh = 0 + \frac{1}{2}g{t^2}\,......\left( {{\text{ii}}} \right) \cr} $$
From $$t = \frac{{nb}}{v},$$   putting in eq. (ii)
$$nh = \frac{1}{2}g \times {\left( {\frac{{nb}}{u}} \right)^2}\,{\text{or}}\,n = \frac{{2h{u^2}}}{{g{b^2}}}$$

For the body starting from rest
$${x_1} = 0 + \frac{1}{2}a{t^2} \Rightarrow {x_1} = \frac{1}{2}a{t^2}$$
For the body moving with constant speed
$$\eqalign{ & {x_2} = vt \cr & \therefore {x_1} - {x_2} = \frac{1}{2}a{t^2} - vt \cr & \Rightarrow \frac{{d\left( {{x_1} - {x_2}} \right)}}{{dt}} = at - v \cr & {\text{at }}t = 0,\,\,\,\,\,\,{x_1} - {x_2} = 0 \cr & {\text{For }}t < \frac{v}{a};\,{\text{the slope is negative}} \cr & {\text{For }}t = \frac{v}{a};\,{\text{the slope is zero}} \cr & {\text{For }}t > \frac{v}{a};\,{\text{the slope is positive}} \cr} $$
These characteristics are represented by graph (B).

$$\eqalign{ & {\text{Given,}}\,\,s = {t^3} - 6{t^2} + 3t + 4 \cr & \therefore {\text{Velocity}}\,v = \frac{{ds}}{{dt}} = 3{t^2} - 12t + 3\,......\left( {\text{i}} \right) \cr & {\text{Acceleration }}a{\text{ is given by}} \cr & a = \frac{{dv}}{{dt}} \cr & \therefore a = 6t - 12\,......\left( {{\text{ii}}} \right) \cr & {\text{For}}\,a = 0,\,{\text{we have}}\,0 = {\text{ }}6t{\text{ }} - 12 \cr & {\text{or}}\,t = 2\,s \cr & {\text{Hence, at }}t{\text{ }} = 2\,s{\text{ the velocity will be}} \cr & v = 3 \times {2^2} - 12 \times 2 + 3 = - 9\,m{s^{ - 1}} \cr} $$

For a body thrown vertically upwards acceleration remains constant ($$a =- g$$  ) and velocity at anytime $$t$$  is given by $$V=u- gt$$
During rise velocity decreases linearly and during fall velocity increases linearly and direction is opposite to each other.
Hence graph (A) correctly depicts velocity versus time.

$$\eqalign{ & \sqrt {{{0.5}^2} + {{0.8}^2} + {c^2}} = 1 \cr & {\text{or}}\,\,c = \sqrt {1 - 0.89} = \sqrt {0.11} \cr} $$

$$\eqalign{ & \tan \beta = \frac{{{v_{{\text{girl}}}} + {v_{{\text{rain}}}}\sin \alpha }}{{{v_{{\text{rain}}}}\cos \alpha }} \cr & \Rightarrow \tan \beta = \tan \beta = \frac{{8 + 10\sin \alpha }}{{10\cos \alpha }} \cr} $$

$$\eqalign{ & x = \frac{1}{{t + 5}} \cr & \therefore v = \frac{{dx}}{{dt}} = \frac{{ - 1}}{{{{\left( {t + 5} \right)}^2}}} \cr & \therefore a = \frac{{{d^2}x}}{{d{t^2}}} = \frac{2}{{{{\left( {t + 5} \right)}^3}}} = 2{x^3} \cr & {\text{Now}}\,\frac{1}{{\left( {t + 5} \right)}} \propto {v^{\frac{1}{2}}} \cr & \therefore \frac{1}{{{{\left( {t + 5} \right)}^3}}} \propto {v^{\frac{3}{2}}} \propto a \cr} $$

$$a = \frac{{{d^2}x}}{{d{t^2}}} = - s\,{\omega ^2}\sin \omega t.$$
On integrating, $$\frac{{dx}}{{dt}} = s\,{\omega ^2}\frac{{\cos \omega t}}{\omega } = s\,\omega \cos \omega t$$
Again on integrating, we get
$$x = s\omega \frac{{\sin \omega t}}{\omega } = s\sin \omega t$$

Velocity at time $$t$$ is $$\tan {45^ \circ } = 1.$$   Velocity at time $$\left( {t = 1} \right)$$  is $$\tan {60^ \circ } = \sqrt 3 .$$    Acceleration is change in velocity in one second $$ = \sqrt 3 - 1.$$