A particle is projected vertically upwards from $O$ with velocity $v$ and a second particle is projected at the same instant from $P$ (at a height h above $O$) with velocity $v$ at an angle of projection $\theta$. The time when the distance between them is minimum is

Q.

A particle is projected vertically upwards from $O$ with velocity $v$ and a second particle is projected at the same instant from $P$ (at a height h above $O$) with velocity $v$ at an angle of projection $\theta$. The time when the distance between them is minimum is

Q.

A particle is projected from point A on the ground at an angle $\theta$ with horizontal. Another particle is projected from point B simultaneously from height 4H above point A with same velocity. A and B are in the same vertical plane, where H is the maximum height attained by the particle A. Both the particles collide at the same time on the ground. Find the angle at which particle B is projected with horizontal:

Q. A particle is projected from ground with velocity $v$ at angle $\theta$ to horizontal. Average velocity of particle between point of projection and maximum height of projectile is-

Q. When a partical is projected from a tower of height $h$ with horizontal with velocity $u$ at an angle $\theta$ above the horizontal, then range of particle (on ground) is

Q. A particle is projected with a speed $V$ from a point O making an angle of ${30}^{\circ }$ with the vertical. At the same instant, a second particle is thrown vertically upwards with a speed $v$ from a point A. The two particles reach $H$, the highest point of the parabolic path of particle one simultaneously. Then ratio $V/v$ is