$A$ ball is projected from the ground with a speed $15 \, m/s$ at an angle $\theta$ with the horizontal such that its range and maximum height are equal. Then,$\tan \theta$ will be equal to:

$A$ ball $P$ is dropped vertically and another ball $Q$ is thrown horizontally with the same initial vertical velocity (which is $0$) from the same height $h$ at the same time. If air resistance is neglected,then:

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

The horizontal range and maximum height attained by a projectile are $R$ and $H$,respectively. If a constant horizontal acceleration $a=g/4$ is imparted to the projectile due to wind,then its horizontal range and maximum height will be

$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight,then the value of maximum range is (acceleration due to gravity $= g$)

The horizontal range $(R)$ of a projectile is $n$ times its maximum height $(H)$. Find the angle of projection $(\theta_0)$.