An object is projected such that it attains maximum range. Another body is projected to reach maximum height. If both objects reach the same maximum height,what is the ratio of their initial velocities?

$A$ ball is thrown upwards at an angle of $60^o$ to the horizontal. It falls on the ground at a distance of $90 \,m$. If the ball is thrown with the same initial velocity at an angle $30^o$,it will fall on the ground at a distance of ........ $m$.

The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$,the range of the projectile (in meters) is:

$A$ particle aimed at a target,projected with an angle $15^{\circ}$ with the horizontal is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$ it is away from the target by $10 \ m$,then the angle of projection to hit the target is

For a projectile,the ratio of the maximum height reached to the square of the time of flight is:

The trajectory of a projectile,projected from the ground,is given by $y = x - \frac{x^2}{20}$,where $x$ and $y$ are measured in meters. The maximum height attained by the projectile will be $...........\,m$.