The angle of projection of a projectile whose path is shown in the given figure is

$A$ body is projected from the ground with a velocity $v = (3 \hat{i} + 10 \hat{j}) \text{ m/s}$. The maximum height attained and the range of the body respectively are (given $g = 10 \text{ m/s}^2$):

Path of projectile is given by the equation $Y = P x - Q x^2$, match the following accordingly (acceleration due to gravity = $g$)

$(A)$ Range	$(i)$ $\frac{P}{Q}$
$(B)$ Maximum height	$(ii)$ $P$
$(C)$ Time of flight	$(iii)$ $\frac{P^2}{4 Q}$
$(D)$ Tangent of projection	$(iv)$ $\left(\sqrt{\frac{2}{g Q}}\right) P$

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:

$A$ particle of mass $m$ is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

$A$ 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 $\,s$)