The equation of motion of a projectile is $y = 12x - \frac{5}{9}x^2$. The horizontal component of velocity is $3 \ ms^{-1}$. Given that $g = 10 \ ms^{-2}$,the range of the projectile is .......... $m$.

$A$ $1.5 \,kg$ ball is shot upward at an angle of $34^{\circ}$ to the horizontal with an initial speed of $20 \,m/s$. The maximum height reached by the ball is (use $\cos 34^{\circ} = 0.83$ and $\sin 34^{\circ} = 0.56$): (in $\,m$)

An aeroplane flying at a constant velocity releases a bomb. As the bomb drops down from the aeroplane,

$A$ cricket ball is thrown at a speed of $28 \; m/s$ in a direction $30^{\circ}$ above the horizontal. Calculate:
$(a)$ the maximum height,
$(b)$ the time taken by the ball to return to the same level,and
$(c)$ the distance from the thrower to the point where the ball returns to the same level.

$A$ body is projected horizontally from the top of a tall tower with a velocity of $30 \,ms^{-1}$. At time $t_1$, its horizontal and vertical components of the velocity are equal and at time $t_2$, its horizontal and vertical displacements are equal. Then $t_2-t_1$ is (take, $g=10 \,ms^{-2}$ ) (in $\,s$)

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: $(g=10 \ m \ s^{-2})$