$A$ 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

$A$ piece of marble is projected from the Earth's surface with a velocity of $19.6 \sqrt{2} \ m/s$ at an angle of $45^{\circ}$. $2 \ s$ later,its velocity makes an angle $\alpha$ with the horizontal,where $\alpha$ is $..........$ (in $^{\circ}$)

If the instantaneous velocity of a particle projected as shown in the figure is given by $v = a \hat{i} + (b - ct) \hat{j}$,where $a, b$,and $c$ are positive constants,the range on the horizontal plane will be

$A$ ball is thrown at a speed of $20 \,m/s$ at an angle of $30^{\circ}$ with the horizontal. The maximum height reached by the ball is (Use $g=10 \,m/s^2$) (in $\,m$)

$A$ projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall of height $H$ at its highest point,as shown in the figure. The horizontal distance from the point of projection to the wall is $\sqrt{3} H$. The angle $\theta$ at which the projectile is thrown is given by:

Two balls with same mass and initial velocity are projected at different angles in such a way that the maximum height reached by the first ball is $8$ times higher than that of the second ball. If $T_1$ and $T_2$ are the total flight times of the first and second ball,respectively,then the ratio of $T_1$ to $T_2$ is: