$A$ projectile is projected in such a way that it achieves maximum range for a given velocity. Find the velocity of the projectile at its maximum height.

$A$ particle is projected from the ground at an angle of $\theta$ with the horizontal with an initial speed of $u$. The time after which the velocity vector of the projectile is perpendicular to the initial velocity is:

If a body is projected with an angle $\theta$ to the horizontal, then

If a stone projected from the ground takes $4 \, s$ to reach the topmost point of its trajectory,then the time of flight is .......... $s$.

$A$ projectile is projected with a velocity of $25 \, m/s$ at an angle $\theta$ with the horizontal. After $t$ seconds,its inclination with the horizontal becomes zero. If $R$ represents the horizontal range of the projectile,the value of $\theta$ will be: [use $g = 10 \, m/s^2$]

$A$ particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterized by