For a projectile,the ratio of maximum height reached to the square of flight time is $(g = 10 \ m/s^2)$.

The range of a particle when launched at an angle of $15^\circ$ with the horizontal is $1.5 \, km$. What is the range of the projectile when launched at an angle of $45^\circ$ to the horizontal? (in $km$)

In a sports event,a disc is thrown such that it reaches its maximum range of $80 \ m$. The distance travelled in the first $3 \ s$ is $(g = 10 \ m/s^2)$. (in $m$)

$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then

The maximum height attained by a projectile is increased by $10\,\%$ by increasing its speed of projection,without changing the angle of projection. The percentage increase in the horizontal range will be $...........\,\%$

$A$ projectile thrown from the ground has initial speed $u$ and its direction makes an angle $\theta$ with the horizontal. If at maximum height from the ground,the speed of the projectile is half its initial speed of projection,then the maximum height reached by the projectile is:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$