$A$ stone is projected in air. Its time of flight is $3\,s$ and range is $150\,m$. The maximum height reached by the stone is $......\,m$ $\left(g=10\,m/s^2\right)$.

$A$ ball having kinetic energy $KE$ is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?

$A$ 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?

$A$ ball is projected from the ground into the air. At a height of $5 \text{ m}$, its velocity is $\vec{v} = (5 \hat{i} + 5 \hat{j}) \text{ m s}^{-1}$. The maximum height reached by the ball is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$): (in $\text{ m}$)

Two paper screens $A$ and $B$ are separated by $150 \ m$. $A$ bullet pierces $A$ and then $B$. The hole in $B$ is $15 \ cm$ below the hole in $A$. If the bullet is traveling horizontally at the time of hitting $A$,then the velocity of the bullet at $A$ is: $(g=10 \ m \ s^{-2})$

$A$ 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. . . . .