$A$ body projected from the ground reaches a point $X$ in its path after $3 \ s$ and from there it reaches the ground after a further $6 \ s$. The vertical distance of the point $X$ from the ground is (acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

$A$ cricketer can throw a ball to a maximum horizontal distance of $100 \; m$. How much high above (in $m$) the ground can the cricketer throw the same ball?

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5 \text{ s}$ and $10 \text{ s}$, respectively, then the value of $R$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

$A$ bomb is released from a horizontal flying aeroplane. The trajectory of the bomb is

$A$ particle is projected from the ground with an initial speed of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is

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: