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Q. A pendulum bob has a speed $3m/s$ while passing through its lowest position, length of the pendulum is $0.5$ $m$ then its speed when it makes an angle of $60^\circ $ with the vertical is :

NTA AbhyasNTA Abhyas 2020

Solution:

Solution
height attained $=\ell \left(\right.1-cosq\left.\right)$
$u^{2}=\sqrt{v^{2} - 2 g \ell \left(\right. 1 - cosq \left.\right)}$