A pendulum bob has a speed of 3 , m/s at its | vedclass

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(D) Let the mass of the bob be $m$, length of the pendulum be $l = 0.5 \, m$, and initial speed at the lowest point be $u = 3 \, m/s$.Using the law of

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(D) Let the mass of the bob be $m$, length of the pendulum be $l = 0.5 \, m$, and initial speed at the lowest point be $u = 3 \, m/s$.Using the law of conservation of mechanical energy between the lowest point $(A)$ and the point where the string makes an angle $	heta = 60^{\circ}$ with the vertical $(B)$:Total energy at $A$ = Total energy at $B$$\frac{1}{2} mu^2 = \frac{1}{2} mv^2 + mgh$Where $h$ is the vertical height gained by the bob, given by $h = l(1 - \cos 	heta)$.Substituting the values:$\frac{1}{2} m(3)^2 = \frac{1}{2} mv^2 + mg(0.5)(1 - \cos 60^{\circ})$$9 = v^2 + 2(10)(0.5)(1 - 0.5)$$9 = v^2 + 10(0.5)$$9 = v^2 + 5$$v^2 = 4$$v = 2 \, m/s$.

$A$ pendulum bob has a speed of $3 \, m/s$ at its lowest position. The pendulum is $50 \, cm$ long. The speed of the bob, when the string makes an angle of $60^{\circ}$ to the vertical, will be $....... \, m/s$ $(g = 10 \, m/s^2)$.

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