$A$ projectile is thrown at an angle $\theta$ with the horizontal and its range is $R_1$. It is then thrown at an angle $\theta$ with the vertical and the range is $R_2$. Then:

$A$ body is projected with a velocity $u$ at an angle $\theta$ with the horizontal. If the time of ascent of the body is $1 \ s$,then the maximum height it can reach is (Take $g = 10 \ m/s^2$) (in $m$)

$A$ ball is thrown upwards at an angle of $60^o$ to the horizontal. It falls on the ground at a distance of $90 \,m$. If the ball is thrown with the same initial velocity at an angle $30^o$,it will fall on the ground at a distance of ........ $m$.

$A$ particle is projected from the ground at an angle of $\theta$ with the horizontal with an initial speed of $u$. The time after which the velocity vector of the projectile is perpendicular to the initial velocity is:

Two particles $A$ and $B$ are projected simultaneously from a fixed point on the ground. Particle $A$ is projected on a smooth horizontal surface with speed $v$,while particle $B$ is projected in air with speed $\frac{2v}{\sqrt{3}}$. If particle $B$ hits particle $A$,the angle of projection of $B$ with the vertical is

$A$ particle $P$ is projected with velocity $u_1$ at an angle of $30^\circ$ with the horizontal. Another particle $Q$ is projected vertically upwards with velocity $u_2$ from the point directly below the maximum height of $P$. The condition for the collision of both particles is: