Speed of walking $$ = \frac{h}{{{t_1}}} = {v_1}$$
Speed of escalator $$ = \frac{h}{{{t_2}}} = {v_2}$$
Time taken when she walks over running escalator
$$\eqalign{ & \Rightarrow t = \frac{h}{{{v_1} + {v_2}}} \cr & \Rightarrow \frac{1}{t} = \frac{{{v_1}}}{h} + \frac{{{v_2}}}{h} = \frac{1}{{{t_1}}} + \frac{1}{{{t_2}}} \cr & \Rightarrow t = \frac{{{t_1}{t_2}}}{{{t_1} + {t_2}}} \cr} $$

$$\eqalign{ & {\bf{Case - 1:}} \cr & u = 50 \times \frac{5}{{18}}\,m/s, \cr & v = 0,\,\,\,\,s = 6\,m,\,\,\,\,a = a \cr & {v^2} - {u^2} = 2as \cr & \Rightarrow {0^2} - {\left( {50 \times \frac{5}{{18}}} \right)^2} = 2 \times a \times 6 \cr & \Rightarrow - {\left( {50 \times \frac{5}{{18}}} \right)^2} = 2 \times a \times 6.....(i) \cr & {\bf{Case - 2:}} \cr & u = 100 \times \frac{5}{{18}}\,m/s, \cr & v = 0,\,\,\,\,s = s,\,\,\,\,a = a \cr & \therefore {v^2} - {u^2} = 2as \cr & \Rightarrow {0^2} - {\left( {100 \times \frac{5}{{18}}} \right)^2} = 2 \times a \times s \cr & \Rightarrow - {\left( {100 \times \frac{5}{{18}}} \right)^2} = 2 \times a \times s.....(ii) \cr & {\text{Dividing (i) and (ii) we get}} \cr & \frac{{100 \times 100}}{{50 \times 50}}{\text{ = }}\frac{{2 \times a \times s}}{{2 \times a \times 6}} \cr & \Rightarrow s = 24\,m \cr} $$

$$\frac{{{H_1}}}{{{H_2}}} = \frac{{\frac{{{u^2}{{\sin }^2}\theta }}{{2g}}}}{{\frac{{{u^2}{{\sin }^2}\left( {{{90}^ \circ } - \theta } \right)}}{{2g}}}} = {\tan ^2}\theta $$

From the properties of vector product, the cross product of any vector with zero is a null vector or zero vector.

$$\eqalign{ & {s_n} = \frac{a}{2}\left( {2n - 1} \right); \cr & {s_{n\, + \,1}} = \frac{a}{2}\left[ {2\left( {n + 1} \right) - 1} \right] = \frac{a}{2}\left( {2n + 1} \right) \cr & \frac{{{s_n}}}{{{s_{n\, + \,1}}}} = \frac{{2n - 1}}{{2n + 1}} \cr} $$

For the given conditions, vectors $$\vec A,\vec B$$  and $${\vec C}$$ are parallel, so $$\vec A \times \vec C = AC\sin {0^ \circ } = 0.$$

For maximum range of projectile, $$\theta $$ will be $${45^ \circ }$$ by the law of projectile motion.
So, maximum range, $${R_{\max }} = \frac{{{u^2}}}{g}$$
Given, $$u = 20\,m{s^{ - 1}}\,{\text{and}}\,g = 10\,m{s^{ - 2}}$$
$$\eqalign{ & {R_{\max }} = \frac{{{{\left( {20} \right)}^2}}}{{10}} = \frac{{400}}{{10}} \cr & {R_{\max }} = 40\,m \cr} $$

In both the cases, the initial velocity in the vertical downward direction is zero. So they will hit the ground simultaneously.

$$\eqalign{ & t = a{x^2} + bx;\,{\text{Different with respect to time }}\left( t \right) \cr & \frac{d}{{dt}}\left( t \right) = a\frac{d}{{dt}}\left( {{x^2}} \right) + b\frac{{dx}}{{dt}} = a.2x\frac{{dx}}{{dt}} + b.\frac{{dx}}{{dt}} \cr & 1 = 2axv + bv = v\left( {2ax + b} \right) \Rightarrow 2ax + b = \frac{1}{v} \cr & {\text{Again differentiating, }}2a\frac{{dx}}{{dt}} + 0 = - \frac{1}{{{v^2}}}\frac{{dv}}{{dt}} \cr & \Rightarrow \frac{{dv}}{{dt}} = f = - 2a{v^3}\,\,\,\,\left( {\because \,\frac{{dv}}{{dt}} = f = acc} \right) \cr} $$

As distance travelled in n^th $$\sec$$  is given by
$${s_n} = u + \frac{1}{2}a\left( {2n - 1} \right)$$
Here, $$u = 0,$$  acceleration due to gravity
$$\eqalign{ & a = 9.8\,m/{s^2} \cr & \therefore {\text{For}}\,\,{4^{th}}s,\,{s_4} = \frac{1}{2} \times 9.8\left( {2 \times 4 - 1} \right) \cr} $$
and for $${5^{th{\text{ }}}}s,\,\,{s_5} = \frac{1}{2} \times 9.8\left( {2 \times 5 - 1} \right)$$
$$\therefore \frac{{{S_4}}}{{{s_5}}} = \frac{7}{9}$$