The equation of a projectile is $y = 16x - \frac{5x^2}{4}$. The horizontal range is .......... $m$.

$A$ ball is projected from a building of height $10 \ m$ with a velocity of $10 \ m/s$ at an angle of $30^\circ$ with the horizontal. What is the horizontal distance covered by the ball when it reaches the same height of $10 \ m$ again (in $m$)? $(g = 10 \ m/s^2, \sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2)$

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 $m$)

The angle of projection of a projectile for which its initial kinetic energy becomes half at its maximum height is (in $^{\circ}$)

Match the columns:

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then: