At the top of the trajectory of a projectile,the directions of its velocity and acceleration are

$A$ body projected at a certain angle $(\neq 90^{\circ})$ from the ground crosses a point in its path at a time $t_1 = 2.3 \ s$ and from there it reaches the ground after a further time $t_2 = 5.7 \ s$. The maximum height reached by the body is (Take $g = 10 \ ms^{-2}$) (in $m$)

If the time of flight of a bullet over a horizontal range $R$ is $T \ s$,the inclination of the direction of projection to the horizontal is

Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.
Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.

$A$ particle is projected from the ground with an initial speed of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is

$A$ ball is thrown at an angle $\theta$ with the horizontal. Its horizontal range is equal to its maximum height. This is possible only when the value of $\tan \theta$ is ..........