The velocity of a projectile at the initial point $A$ is $(2\hat i + 3\hat j) \text{ m/s}$. Its velocity (in $\text{m/s}$) at point $B$ is

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	$(iv)$ $\left(\sqrt{\frac{2}{g Q}}\right) P$

$A$ projectile is thrown at an angle of $\pi/3$ with the horizontal,reaching a maximum height $Y$. If another projectile is thrown with the same initial velocity at an angle of $\pi/6$,what will be the maximum height attained?

$A$ ball is projected from the ground with a velocity $V$ at an angle $\theta$ to the vertical. On its path,it makes an elastic collision with a vertical wall and returns to the ground. The total time of flight of the ball is:

Assume that the acceleration due to gravity on the surface of the moon is $0.2$ times the acceleration due to gravity on the surface of the earth. If $R_e$ is the maximum range of a projectile on the earth's surface,what is the maximum range on the surface of the moon for the same velocity of projection (in $,R_e$)?

$A$ ball is projected from the ground with a speed $15 \, m/s$ at an angle $\theta$ with the horizontal such that its range and maximum height are equal. Then,$\tan \theta$ will be equal to: