$A$ cricket ball is hit at $30^{\circ}$ with the horizontal with kinetic energy $K$. The kinetic energy at the highest point is

For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then:

$A$ bullet is fired with a velocity $u$ making an angle of $60^{\circ}$ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is

$A$ projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$,the range of the projectile is $d$. If at the highest point in its trajectory,the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$,then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

Two particles are simultaneously projected in opposite directions horizontally from a given point in space,where gravity $g$ is uniform. If $u_1$ and $u_2$ are their initial speeds,then the time $t$ after which their velocities are mutually perpendicular is given by

$A$ cricket ball is thrown at a speed of $28 \; m/s$ in a direction $30^{\circ}$ above the horizontal. Calculate:
$(a)$ the maximum height,
$(b)$ the time taken by the ball to return to the same level,and
$(c)$ the distance from the thrower to the point where the ball returns to the same level.