$A$ particle is thrown with a speed of $12 \, m/s$ at an angle $60^o$ with the horizontal. The time interval between the moments when its speed is $10 \, m/s$ is $(g = 10 \, m/s^2)$.........$s$

$A$ 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:

For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then:

$A$ ball is thrown from a point with a speed $v_0$ at an angle of projection $\theta$. From the same point and at the same instant,a person starts running with a constant speed $v_0/2$ to catch the ball. Will the person be able to catch the ball? If yes,what should be the angle of projection?

$A$ ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \ m$ high and $0.1 \ m$ wide. The minimum velocity $u$ with which the ball just hits the $5^{\text{th}}$ step of the stairway will be $\sqrt{x} \ ms^{-1}$ where $x=$ . . . . . . [use $g=10 \ ms^{-2}$].

The horizontal range and maximum height attained by a projectile are $R$ and $H$,respectively. If a constant horizontal acceleration $a=g/4$ is imparted to the projectile due to wind,then its horizontal range and maximum height will be