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(A) Let the particle be at point $A$ (horizontal position) and move to point $B$ at an angle $	heta = 30^o$ with the horizontal. The height $h$ gaine

Vedclass · Accepted Answer

$\sqrt{lg}$

Vedclass · Answer

(A) Let the particle be at point $A$ (horizontal position) and move to point $B$ at an angle $	heta = 30^o$ with the horizontal. The height $h$ gained by the particle is $l \sin 30^o = \frac{l}{2}$.By the work-energy theorem, the kinetic energy at $B$ is given by $\frac{1}{2}mv_B^2 = \frac{1}{2}mu^2 - mgh = \frac{1}{2}mu^2 - mg(\frac{l}{2})$. Thus, $v_B^2 = u^2 - gl$.The tension $T$ at point $B$ is given by $T - mg \sin 30^o = \frac{mv_B^2}{l}$.Given $T = \frac{mg}{2}$, we have $\frac{mg}{2} - mg \sin 30^o = \frac{mv_B^2}{l}$.Since $\sin 30^o = \frac{1}{2}$, we get $\frac{mg}{2} - \frac{mg}{2} = \frac{mv_B^2}{l}$, which implies $v_B = 0$.Substituting $v_B = 0$ into the energy equation: $0 = u^2 - gl$, so $u^2 = gl$.Therefore, $u = \sqrt{lg}$.

$A$ 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

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