Time taken by the body to reach the ground from some height is the same as taken to reach that height. Hence, time to reach the ground from its maximum height is $$5\,s.$$

$${\text{Average acceleration}} = \frac{{{\text{change in velocity}}}}{{{\text{time interval}}}}$$

$$\eqalign{ & = \frac{{\Delta \overrightarrow v }}{t} \cr & \overrightarrow {{v_1}} = 5\hat i,\,\,\overrightarrow {{v_2}} = 5\hat j \cr & \therefore \overrightarrow a = \frac{{5\hat j - 5\hat i}}{{10}} = \frac{{\hat j - \hat i}}{2} \cr & \therefore a = \frac{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} }}{2} = \frac{{\sqrt 2 }}{2} = \frac{1}{{\sqrt 2 }}\,m{s^{ - 2}} \cr & \tan \theta = \frac{{{v_2}}}{{{v_1}}} = \frac{5}{5} = 1 \cr & \therefore \theta = {45^ \circ } \cr} $$
Therefore the direction is North-west.

From figure,
$$\eqalign{ & \overrightarrow {OA} = 0\hat i + 30\hat j, \cr & \overrightarrow {AB} = 20\hat i + 0\vec j, \cr & \overrightarrow {BC} = - 30\sqrt 2 \cos {45^ \circ }\hat i - 30\sqrt 2 \cos {45^ \circ }\hat j \cr & = - 30\hat i - 30\vec j \cr} $$
$$\therefore $$ Net displacement,
$$\eqalign{ & \overrightarrow {OC} = \overrightarrow {OA} + \overrightarrow {AB} + \overrightarrow {BC} = - 10\hat i + 0\hat j \cr & = \left| {\overrightarrow {OC} } \right| = 10\,m. \cr} $$

$$\eqalign{ & S = ut + \frac{1}{2}a{t^2} \cr & - 76 = 4 \times 6 - \frac{1}{2} \times 10 \times {\left( 6 \right)^2} \Rightarrow u = \frac{{52}}{3}m/s \cr} $$

Change in velocity = area under the graph $$ = \frac{1}{2} \times 10 \times 11 = 55\,m/s$$

Since, initial velocity is zero, final velocity is 55 m/s.

Velocity of projectile $$u = 147\,m{s^{ - 1}}$$
angle of projection $$\alpha = {60^ \circ }$$
Let, the time taken by the Projectile from $$O$$ to $$A$$ be $$t$$ where direction $$\beta = {45^ \circ }.$$  As horizontal component of velocity remains constant during the projectile motion.

$$\eqalign{ & \Rightarrow v\cos {45^ \circ } = u\cos {60^ \circ } \cr & \Rightarrow v \times \frac{1}{{\sqrt 2 }} = 147 \times \frac{1}{2}\, \Rightarrow v = \frac{{147}}{{\sqrt 2 }}m{s^{ - 1}} \cr} $$
For Vertical motion,
$$\eqalign{ & {v_y} = {u_y} - gt \cr & \Rightarrow v\sin {45^ \circ } = 45\sin {60^ \circ } - 9.8t \cr & \Rightarrow \frac{{147}}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} = 147 \times \frac{{\sqrt 3 }}{2} - 9.8t \cr & \Rightarrow 9.8t = \frac{{147}}{2}\left( {\sqrt 3 - 1} \right) \Rightarrow t = 5.49\,s \cr} $$

When string makes an angle $$\theta $$ with the vertical in a vertical circle, then balancing the force we get
$$T - mg\cos \theta = \frac{{m{v^2}}}{l}$$

$${\text{or}}\,T = mg\cos \theta + \frac{{m{v^2}}}{l}$$
Tension is maximum when $$\cos \theta = + 1$$
i.e. $$\theta = 0$$
Thus, $$\theta $$ is zero at lowest point $$B.$$ At this point tension is maximum. So, string will break at point $$B.$$
NOTE
The critical speed of a body on circular path, $${v_c} = \sqrt {Rg} ,R = $$   radius of path.
If at the highest point, the speed is less than this, the string would become slack and the body would leave the circular path.

Let the components of $$\overrightarrow A $$ make angles $$\alpha ,\beta $$  and $$\gamma $$ with $$x,y$$  and $$z$$-axis respectively then $$\alpha = \beta = \gamma $$   and
$$\eqalign{ & {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \cr & \Rightarrow 3{\cos ^2}\alpha = 1 \Rightarrow \cos \alpha = \frac{1}{{\sqrt 3 }} \cr & \therefore {A_x} = {A_y} = {A_z} = A\cos \alpha = \frac{A}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & {v^2} = {u^2} - 2gh \cr & {\text{or}},\,v = \sqrt {{u^2} - 2gh} \cr & {\text{Momentum, }}\,p = mv{\text{ }} \cr & \therefore \,p = m\sqrt {{u^2} - 2gh} \cr} $$
Therefore graph between $$p$$  and $$h$$ cannot have straight line.
(B) and (C) are not possible.
During upward journey as $$h$$  increases, $$p$$  decreases and in downward journey as $$h$$  decreases $$p$$  increases.
Therefore (D) is the correct option.

$$\eqalign{ & \vec v = k\left( {y\,\hat i + x\,\hat j} \right) = {v_x}\hat i + {v_y}\hat j = \frac{{dx}}{{dt}}\hat i + \frac{{dy}}{{dt}}\hat j \cr & \therefore \frac{{dx}}{{dt}} = ky\,\,{\text{and }}\therefore \frac{{dy}}{{dt}} = kx \cr & \therefore \frac{{dy}}{{dx}} = \frac{x}{y} \cr & \Rightarrow y\,dy = x\,dx \cr & \Rightarrow {y^2} = {x^2} + \,\,{\text{constant}} \cr} $$