$A$ 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 $= g$)

$A$ projectile is fired at $30^{\circ}$ to the horizontal. The vertical component of its velocity is $80 \; ms^{-1}$. Its time of flight is $T$. What will be the velocity of the projectile at $t = \frac{T}{2}$?

The equation of a projectile is $y = \sqrt{3} x - \frac{gx^2}{2}$. The angle of projection is:

For a projectile motion,the angle between the velocity and acceleration is minimum and acute at

$A$ projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$,the range of the projectile is $d$. If at the highest point in its trajectory,the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$,then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

When a body is thrown with a velocity $u$ making an angle $\theta$ with the horizontal plane,the maximum distance covered by it in the horizontal direction is: