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

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

A bowling machine placed at a height $h$ above the earth surface releases different balls with different angles but with same velocity $10 \sqrt{3} \mathrm{~ms}^{-1}$. All these balls landing velocities make angels $30^{\circ}$ or more with horizontal. Then the height $h$ (in metre) (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A boy throws a ball with a velocity $v_0$ at an angle $\alpha$ to the ground. At the same time he starts running with uniform velocity to catch the ball before it hits the ground. To achieve this, he should run with a velocity of

A ball at point $O$ is at a horizontal distance of 7 m from a wall. On the wall a target is set at point $C$. If the ball is throw from $O$ at an angle $37^{\circ}$ with horizontal aiming the target $C$. But it hits the wall at point $D$ which is at a vertical distance $y_0$ below $C$. If the initial velocity of the ball is $15 \mathrm{~ms}^{-1}$. Find $y_0\left(\right.$ given, $\left.\cos 37^{\circ}=\frac{4}{5}\right)$