$A$ wheel of radius $R$ is trapped in a mud pit and spinning. As the wheel is spinning,it splashes mud blobs with initial speed $u$ from various points on its circumference. The maximum height from the centre of the wheel,to which a mud blob can reach is

$A$ particle is projected from the ground with velocity $u$ at an angle $\theta$ with the horizontal. The horizontal range,maximum height,and time of flight are $R, H$,and $T$ respectively. They are given by $R = \frac{u^2 \sin 2\theta}{g}$,$H = \frac{u^2 \sin^2 \theta}{2g}$,and $T = \frac{2u \sin \theta}{g}$. Now,keeping $u$ fixed,$\theta$ is varied from $30^o$ to $60^o$. Then,

$A$ ball is hit by a batsman at an angle of $37^o$ with an initial velocity $u = 15 \ m/s$ as shown in the figure. The man standing at $P$ is at a distance of $9 \ m$ from the point $B$ where the ball would strike the ground. What is the minimum velocity at which the man should run to catch the ball before it hits the ground? Assume the height of the man is negligible. ........ $m/s$

Fill in the blanks:
$(a)$ If $\overrightarrow A \cdot \overrightarrow B = AB$,then the angle between $\overrightarrow A$ and $\overrightarrow B$ is ............
$(b)$ The velocity of a projectile at its maximum height is ............ (Take the angle of projection as $\theta$).
$(c)$ The projection of $\widehat i - 2\widehat j + 4\widehat k$ on the $y$-axis is ............

$A$ ball of mass $1 \; kg$ is thrown vertically upwards and returns to the ground after $3 \; s$. Another ball,thrown at $60^{\circ}$ with the vertical,also stays in the air for the same time before it touches the ground. The ratio of the two maximum heights reached is:

An object of mass $m$ is projected with a momentum $p$ at such an angle that its maximum height is $\frac{1}{4}$th of its horizontal range. Its minimum kinetic energy in its path will be