$A$ projectile is thrown with a velocity of $50 \, ms^{-1}$ at an angle of $53^o$ with the horizontal. The equation of the trajectory is given by:

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

$A$ ball can be thrown to a maximum horizontal distance of $100\,m$. To what maximum height can it be thrown?

$A$ ball is thrown from a point with a speed $v_0$ at an elevation angle of $\theta$. From the same point and at the same instant,a person starts running with a constant speed $\frac{v_0}{2}$ to catch the ball. Will the person be able to catch the ball? If yes,what should be the angle of projection $\theta$?

$A$ 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).

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