$A$ stone is projected from the ground with velocity $25\,m/s$. Two seconds later,it just clears a wall $5\,m$ high. The angle of projection of the stone is ........ $^o$ $(g = 10\,m/s^2)$.

$A$ ball thrown by one player reaches another player in $2 \text{ s}$. The maximum height attained by the ball above the point of projection will be $.... \text{ m}$.

$A$ body when projected at an angle $\theta$ with the horizontal reaches a maximum height $H$. The time of flight of the body will be ($g=$ acceleration due to gravity).

The speed of a projectile at its maximum height is $\frac{\sqrt{3}}{2}$ times its initial speed. If the range of the projectile is $P$ times the maximum height attained by it,$P$ is equal to

$A$ cricket player hits a ball like a projectile,and the fielder catches the ball after $2 \ s$. The maximum height reached by the ball is $(g = 10 \ m/s^2)$. (in $m$)

The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)