A ball is projected from a point with a speed $V_0$ at certain angle with the horizontal. From the same point and at the same instant, a person starts running with a constant speed $0.5 V_0$ to catch the ball. If the person catches the ball after some time, then the angle of projection of the ball is

Path of projectile is given by the equation $Y=P x-Q x^2$, match the following accordingly (acceleration due to gravity $=g$ )

$$ \begin{array}{llll} \hline \text { a. } & \text { Range } & \text { i } & \frac{P}{Q} \\ \hline \text { b. } & \text { Maximum height } & \text { ii } & P \\ \hline \text { c. } & \text { Time of flight } & \text { iii } & \frac{P^2}{4 Q} \\ \hline \text { d. } & \text { Tangent of projection } & \text { iv } & \left(\sqrt{\frac{2}{g Q}}\right) P \\ \hline \end{array} $$