If we can throw a ball up to a maximum height $H$,the maximum horizontal distance to which we can throw it is

$A$ 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.

Four objects $P, Q, R,$ and $S$ are projected with the same velocity at angles of $15^o, 30^o, 45^o,$ and $60^o$ respectively. Which of these will have the minimum range?

An object is projected in the air with initial velocity $u$ at an angle $\theta$. The projectile motion is such that the horizontal range $R$ is maximum. Another object is projected in the air with a horizontal range half of the range of the first object. The initial velocity remains the same in both cases. The value of the angle of projection,at which the second object is projected,will be $.......$ degrees.

The path of a projectile is given by the equation $y = ax - bx^2$, where $a$ and $b$ are constants, and $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection, respectively. The maximum height attained by the projectile and the angle of projection are respectively:

$A$ particle is projected with a velocity of $30\,m/s$ at an angle of $\theta_0 = \tan^{-1}(3/4)$. After $1\,s$,the particle is moving at an angle $\theta$ to the horizontal. What is the value of $\tan\theta$? (Take $g = 10\,m/s^2$)