$$\eqalign{ & y = x\tan \theta - \frac{1}{2}\frac{{g{x^2}}}{{{u^2}{{\cos }^2}\theta }} \cr & y = 50\tan {60^ \circ } - \frac{{10 \times 50 \times 50}}{{2 \times 25 \times 25 \times {{\cos }^2}{{60}^ \circ }}} = 5\,m \cr} $$

Time of flight $$ = \frac{{2u\sin \theta }}{g}$$
$$\eqalign{ & = \frac{{2 \times 9.8 \times \sin {{30}^ \circ }}}{{9.8}} \cr & = 2 \times \frac{1}{2} \cr & = 1\,\sec . \cr} $$

Let the initial velocity of ball be $$u$$
$$\therefore $$ Time of rise $${t_1} = \frac{u}{{g + a}}$$  and height reached $$ = \frac{{{u^2}}}{{2\left( {g + a} \right)}}$$
Time of fall $${t_2}$$ is given by
$$\eqalign{ & \frac{1}{2}\left( {g - a} \right)t_2^2 = \frac{{{u^2}}}{{2\left( {g + a} \right)}} \cr & {t_2} = \frac{u}{{\sqrt {\left( {g + a} \right)\left( {g - a} \right)} }} \cr & = \frac{u}{{\left( {g + a} \right)}}\sqrt {\frac{{g + a}}{{g - a}}} \cr & \therefore {t_2} > {t_1}\,{\text{because}}\,\frac{1}{{g + a}} < \frac{1}{{g - a}} \cr} $$

$$\eqalign{ & \vec r = 2{t^2}\hat i + 3t\hat j + 4\hat k \cr & \therefore \vec v = \frac{{d\vec r}}{{dt}} = \frac{d}{{dt}} = \left( {2{t^2}\hat i + 3t\hat j + 4\hat k} \right) = 4t\hat i + 3\hat j \cr & \vec a = \frac{{d\vec v}}{{dt}} = \frac{d}{{dt}}\left( {4t\hat i + 3\hat j} \right) = 4\hat i \cr} $$
$$\therefore \vec a = 4\,m{s^{ - 2}}$$   along $$x$$-direction

Given, that two stones of masses $$m$$ and $$2\,m$$  are whirled in horizontal circles, the heavier one in a radius $$\frac{r}{2}$$ and lighter one in radius $$r$$ as shown in figure.

As, lighter stone is $$n$$ times that of the value of heavier stone when they experience same centripetal forces, we get
$$\eqalign{ & {\left( {{F_c}} \right)_{{\text{heavier}}}} = {\left( {{F_c}} \right)_{{\text{lighter}}}} \cr & \Rightarrow \frac{{2m{{\left( v \right)}^2}}}{{\left( {\frac{r}{2}} \right)}} = \frac{{m{{\left( {nv} \right)}^2}}}{r} \cr & \Rightarrow {n^2} = 4 \Rightarrow n = 2 \cr} $$

Required distance from the wall $$ = \frac{{{u^2}\sin 2\theta }}{g} - u\cos \theta \times t = 50\sqrt 3 \,m$$

So by figure the velocity of parrot w.r.t. train is $$ = 5 - \left( { - 10} \right) = 15\,m/\sec $$
so time taken to cross the train is $$ = \frac{{{\text{length of train}}}}{{{\text{relative velocity}}}} = \frac{{150}}{{15}} = 10\,\sec $$

Let $$v$$ be the relative velocity of scooter $$\left( s \right)$$  w.r.t. bus $$\left( B \right),$$  then
$$v = {v_s} - {v_B}$$

$$\therefore {v_s} = v + {v_B}\,......\left( {\text{i}} \right)$$
$${\text{Relative velocity = }}\frac{{{\text{Displacement}}}}{{{\text{Time}}}}$$
$$ = \frac{{1000}}{{100}} = 10\,m{s^{ - 1}}$$
Now, substituting the value of $$v$$ in Eq. (i), we get
$${v_s} = 10 + 10 = 20\,m{s^{ - 1}}$$

According to conservation of energy, the kinetic energy of car = work done in stopping the car i.e. $$\frac{1}{2}m{v^2} = Fs$$
where, $$F$$ is the retarding force and $$s$$ is the stopping distance.
For same retarding force,
$$\eqalign{ & s \propto {v^2} \cr & \therefore \frac{{{s_2}}}{{{s_1}}} = {\left( {\frac{{{v_2}}}{{{v_1}}}} \right)^2} = {\left( {\frac{{80}}{{40}}} \right)^2} = 4 \cr & \therefore {s_2} = 4{s_1} = 4 \times 2 \cr & = 8\,m \cr} $$
Alternative
Initial speed of car $$u = 40\;km/h$$
$$ = 40 \times \frac{5}{{18}}\;m/s = \frac{{100}}{9}\;m/s$$
From 3rd equation of motion,
$$\eqalign{ & {v^2} = {u^2} - 2as \cr & \Rightarrow 0 = {\left( {\frac{{100}}{9}} \right)^2} - 2 \times a \times 2 \cr & 4a = \frac{{100 \times 100}}{{81}} \cr & \Rightarrow a = \frac{{2500}}{{81}}\;m/{s^2} \cr} $$
Final speed of car $$ = 80\;km/h$$
$$ = 80 \times \frac{5}{{18}} = \frac{{200}}{9}\;m/s$$
Suppose car stops for a distance $${s'}.$$ Then
$$\eqalign{ & {v^2} = {u^2} - 2as' \cr & 0 = {\left( {\frac{{200}}{9}} \right)^2} - 2 \times \frac{{2500}}{{81}}s' \cr & \Rightarrow s' = \frac{{200 \times 200 \times 81}}{{9 \times 9 \times 2 \times 2500}} = 8\;m \cr} $$

$$H = \frac{{u_ \bot ^2}}{{2{g_ \bot }}} \Rightarrow H = \frac{{10\sin {{37}^ \circ } \times 10\sin {{37}^ \circ }}}{{2 \times g\cos {{53}^ \circ }}} = 3\,m$$