The trajectory of a projectile projected at some angle $$\theta $$ with the horizontal direction from ground is given by
$$y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$$
For equal trajectories for same angle of projection,
$$\eqalign{ & \frac{g}{{{u^2}}} = {\text{constant}} \Rightarrow \frac{{9.8}}{{{5^2}}} = \frac{{g'}}{{{3^2}}} \cr & g' = \frac{{9.8 \times 9}}{{25}} = 3.5\,m{s^{ - 2}} \cr} $$

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

$$\eqalign{ & \Rightarrow S + f{t_1}t + 2S = 15\,S \cr & \Rightarrow f{t_1}t = 12\,S\,......\left( {\text{i}} \right) \cr & \frac{1}{2}ft_1^2 = S\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii), we get $${t_1} = \frac{t}{6}$$
$$ \Rightarrow S = \frac{1}{2}f{\left( {\frac{t}{6}} \right)^2} = \frac{{f{t^2}}}{{72}}$$

$$\eqalign{ & h = u{t_1} - \frac{1}{2}gt_1^2 \cr & {\text{Also}}\,h = u{t_2} - \frac{1}{2}gt_2^2 \cr} $$
After simplify above equations, we get
$$h = \frac{1}{2}g{t_1}{t_2}.$$

$$\frac{{dx}}{{dt}} = \frac{{{a_1}}}{2} + \frac{{2{a_2}t}}{3} \Rightarrow \frac{{{d^2}x}}{{d{t^2}}} = \frac{{2{a_2}}}{3}$$
acceleration required

Maximum possible horizontal range $$ = \frac{{{v^2}}}{g}$$
Maximum possible area of the circle $$ = \pi {\left( {\frac{{{v^2}}}{g}} \right)^2} = \frac{{\pi {v^4}}}{{{g^2}}}\,\,\left[ {{\text{Here}}\,r = \frac{{{v^2}}}{g}} \right]$$

$${v_L} = 8\,km/h,s = {v_0} \times t \Rightarrow t = \frac{{40}}{8} = 5h.$$
Total distance flown by the bird $$ = 10 \times 5 = 50\,km.$$

$$\eqalign{ & v = 4t - 3{t^2} \cr & \frac{{ds}}{{dt}} = 4t - 3{t^2};\int\limits_0^s {ds} = \int\limits_0^t {\left( {4t - 3{t^2}} \right)dt} \cr & s = \frac{{4{t^2}}}{2} - \frac{{3{t^3}}}{3};s = 2{t^2} - {t^3} \cr & {v_{avg}} = \frac{{{s_{\left( 2 \right)}} - {s_{\left( 0 \right)}}}}{{2 - 0}}\,\left[ {{s_{\left( 2 \right)}} = 8 - 8 = 0} \right] \cr & {v_{avg}} = \frac{{0 - 0}}{2} = 0 \cr} $$

$$\eqalign{ & {x_1} = \frac{1}{2}a{\left( {10} \right)^2} = 50\,a \cr & {x_2} = \left( {10\,a} \right)\left( {10} \right) + \frac{1}{2}a{\left( {10} \right)^2} = 150\,a \cr & {x_3} = \left( {20\,a} \right)\left( {10} \right) + \frac{1}{2}a{\left( {10} \right)^2} = 250\,a \cr & \Rightarrow {x_1}:{x_2}:{x_3}::1:3:5 \cr} $$

From equation, $$\vec v = \hat i + 2\hat j$$
$$\eqalign{ & \Rightarrow x = t\,......\left( {\text{i}} \right) \cr & y = 2t - \frac{1}{2}\left( {10{t^2}} \right)\,......\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii),
$$y = 2x - 5{x^2}$$

$$\eqalign{ & \left| {{\text{Average velocity}}} \right| = \frac{{\left| {{\text{displacement}}} \right|}}{{{\text{time}}}} \cr & = \frac{{2r}}{t} = 2 \times \frac{1}{1} = 2\,m/s. \cr} $$