$A$ particle of mass $m$ is tied to one end of a string of length $l$. The particle is held horizontal with the string taut. It is then projected upward with a velocity $u$. The tension in the string is $\frac{mg}{2}$ when it is inclined at an angle $30^o$ to the horizontal. The value of $u$ is

$A$ body tied to a string of length $L$ is revolved in a vertical circle with minimum velocity. When the body reaches the uppermost point,the string breaks and the body moves under the influence of the gravitational field of the Earth along a parabolic path. The horizontal range $AC$ of the body will be

$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the highest and lowest positions of the circular path will be:

$A$ bullet of mass $m$ and velocity $v$ strikes the bob of a pendulum of mass $M$ and emerges out with velocity $v/2$. What is the minimum value of $v$ such that the bob of the pendulum will swing through a complete circle? (Take the length of the pendulum $= \ell$)

$A$ simple pendulum with a bob of mass $m = 1 \ kg$, charge $q = 5 \ \mu C$ and string length $l = 1 \ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 \times 10^6 \ V/m$ at its bottom-most point $A$, as shown in the figure. It is given a speed $u$ such that the particle leaves the circular path at its topmost point $C$. Find the speed $u$. (Take $g = 10 \ m/s^2$)

$A$ particle of mass $m$,attached to a string of length $l$,is moving in a vertical circle. If the particle is just looping the loop without the string slackening,and $v_A, v_B, v_D$ are the speeds at positions $A, B, D$ shown in the figure,then: