For free fall from a height, $$u = 0$$
$$\therefore $$ Distance covered by stone in first $$5 s,$$
$${h_1} = 0 + \frac{1}{2}g{\left( 5 \right)^2} = \frac{{25}}{2}g\,......\left( {\text{i}} \right)$$
$$\therefore $$ Distance covered in first $$10 s,$$
$${s_2} = 0 + \frac{1}{2}g{\left( {10} \right)^2} = \frac{{100}}{2}g$$
$$\therefore $$ Distance covered in second $$5 s$$
$${h_2} = {s_2} - {h_1} = \frac{{100}}{2}g - \frac{{25}}{2}g = \frac{{72}}{2}g\,......\left( {{\text{ii}}} \right)$$
$$\therefore $$ Distance covered in first $$15 s,$$
$${s_3} = 0 + \frac{1}{2}g{\left( {15} \right)^2} = \frac{{225}}{2}g$$
$$\therefore $$ Distance coveted in last $$5 s,$$
$${h_3} = {s_3} - {s_2} = \frac{{225}}{2}g - \frac{{100}}{2}g = \frac{{125}}{2}g\,......\left( {{\text{iii}}} \right)$$
From Eqs. (i), (ii) and (iii), we get
$$\eqalign{ & {h_1}:{h_2}:{h_3} = \frac{{25}}{2}g:\frac{{75}}{2}g:\frac{{125}}{2}g = 1:3:5 \cr & \Rightarrow {h_1} = \frac{{{h_2}}}{3} = \frac{{{h_3}}}{5} \cr} $$

Let the total distance be $$d.$$ Then for first half distance, time $$ = \frac{d}{{2{v_0}}},$$   next distance. $$ = {v_1}t$$  and last half distance $$ = {v_2}t$$
$$\therefore {v_1}t + {v_2}t = \frac{d}{2};t = \frac{d}{{2\left( {{v_1} + {v_2}} \right)}}$$
Now average speed
$$\eqalign{ & t = \frac{d}{{\frac{d}{{2{v_0}}} + \frac{d}{{2\left( {{v_1} + {v_2}} \right)}} + \frac{d}{{2\left( {{v_1} + {v_2}} \right)}}}} \cr & = \frac{{2{v_0}\left( {{v_1} + {v_2}} \right)}}{{\left( {2{v_0} + {v_1} + {v_2}} \right)}} \cr} $$

We know that
$$s = ut + \frac{1}{2}g{t^2},\,\,or\,h = \frac{1}{2}g{T^2}\left( {\because u = 0} \right)$$
now for $$\frac{T}{3}$$ second, vertical distance moved is given by
$$h' = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} \Rightarrow h' = \frac{1}{2} \times \frac{{g{T^2}}}{9} = \frac{h}{9}$$
$$\therefore $$ position of ball from ground $$ = h - \frac{h}{9} = \frac{{8h}}{9}$$

$$\Delta v = \sqrt 2 v = \sqrt 2 \omega r = \sqrt 2 \left( {\frac{{2\pi }}{{60}}} \right) \times 1 = \frac{{\pi \sqrt 2 }}{{30}}cm/s$$

distance from $$A$$ to $$B$$ $$ = S = \frac{1}{2}ft_1^2 \Rightarrow ft_1^2 = 2S$$
Distance from $$B$$ to $$C$$ $$ = \left( {f{t_1}} \right)t$$
distance from $$C$$ to $$D$$ $$ = \frac{{{u^2}}}{{2a}} = \frac{{{{\left( {f{t_1}} \right)}^2}}}{{2\left( {\frac{f}{2}} \right)}} = ft_1^2 = 2S$$

$$\eqalign{ & \Rightarrow S + f\,{t_1}t + 2\,S = 15\,S \cr & \Rightarrow f\,{t_1}t = 12\,S \cr & {\text{But}}\,\frac{1}{2}f\,t_1^2 = S \cr} $$
On dividing the above two equations, we get $${t_1} = \frac{t}{6}$$
$$ \Rightarrow S = \frac{1}{2}f{\left( {\frac{t}{6}} \right)^2} = \frac{{f\,{t^2}}}{{72}} = \frac{1}{{72}}f\,{t^2}$$

$$\eqalign{ & {\text{Clearly }}\vec a = {a_c}\cos \,\theta \left( { - \hat i} \right) + {a_c}\sin \,\theta \left( { - \hat j} \right) \cr & = \frac{{ - {v^2}}}{R}\cos \theta \hat i - \frac{{{v^2}}}{R}\sin \theta \hat j \cr} $$

Let the velocity along $$x$$ and $$y$$-axes be $${v_x}$$ and $${v_y}$$ respectively.
$$\therefore {v_x} = \frac{{dx}}{{dt}}\,{\text{and}}\,{v_y} = \frac{{dy}}{{dt}}$$
From figure,
$$\tan \alpha = \frac{y}{x} \Rightarrow y = x\tan \alpha $$
Differentiating Eq. (i), w.r.t. $$t,$$ we get
$$\eqalign{ & \frac{{dy}}{{dt}} = \frac{{dx}}{{dt}}\tan \alpha \cr & \Rightarrow \quad {v_y} = {v_x}\tan \alpha \cr & {\text{Here,}}\,\,{v_x} = 10\;m/s,\alpha = {60^ \circ } \cr & \therefore \quad {v_y} = 10\tan {60^ \circ } \cr & = 10\sqrt 3 = 17.3\;m/s \cr} $$

When an object falls freely under gravity, then its speed depends only on its height of fall and is independent of the mass of the object. As all objects are falling through the same height, therefore their speeds on reaching the ground will be in the ratio of $$1:1:1.$$
Alternative
The vertical displacement for all the three is same and paths are frictionless. So, by conservation of mechanical energy,
$$\eqalign{ & \frac{1}{2}m{v^2} = mgl \Rightarrow v = \sqrt {2gl} \cr & {\text{So,}}\,{v_1}:{v_2}:{v_3} = 1:1:1 \cr} $$

As $$\Delta LOS$$  and $$\Delta MPS$$  are similar so,
$$\eqalign{ & \frac{H}{x} = \frac{h}{{x - y}} \Rightarrow \left( {H - h} \right)x = Hy \cr & \Rightarrow \left( {H - h} \right)\frac{{dx}}{{dt}} = H\frac{{dy}}{{dt}} = Hv \cr & \Rightarrow {V_s} = \frac{{Hv}}{{H - h}} \cr} $$

First $$X$$ by $$X$$ differentiating displacement equation we get velocity of the body, since body comes to rest so velocity becomes zero. Now by putting the value of time $$t$$ in displacement equation we get the distance travelled by the body when it comes to rest.
Distance travelled by the particle is $$x = 40 + 12t - {t^3}$$
We know that, velocity is the rate of change of distance i.e. $$v = \frac{{dx}}{{dt}}.$$
$$\eqalign{ & \therefore v = \frac{d}{{dt}}\left( {40 + 12t - {t^3}} \right) \cr & = 0 + 12 - 3{t^2} \cr} $$
but final velocity $$v = 0$$
$$\eqalign{ & \therefore 12 - 3{t^2} = 0 \cr & {\text{or}}\,{t^2} = \frac{{12}}{3} = 4 \cr & {\text{or}}\,t = 2\,s \cr} $$
Hence, distance travelled by the particle before coming to rest is given by
$$\eqalign{ & x = 40 + 12\left( 2 \right) - {\left( 2 \right)^3} \cr & = 40 + 24 - 8 \cr & = 64 - 8 \cr & = 56\,m \cr} $$