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The bob of a pendulum is released from a horizontal position. If the length of the pendulum is $1.5 \mathrm{~m}$, what is the speed with which the bob arrives at the lowermost point, given that it dissipated $5 \%$ of its initial energy against air resistance?

Length of the pendulum is given as $\mid=1.5 \mathrm{~m}$

Potential of the bob at the horizontal position is given as $=m g h=m g \mid$

When the bob goes from the horizontal position to the lowest point, the initial energy dissipated against air resistance is given as $5 \%$

The total kinetic energy of the bob at the lowermost position is $95 \%$ of the total potential energy at the horizontal position

$(1 / 2) m v^{2}=(95 / 100) \mathrm{mgl}$

$v^{2}=2[(95 / 100) \times 9.8 \times 1.5]$

$v^{2}=2(13.965)=27.93$

$v=\sqrt{27.93}=5.28 \mathrm{~m} / \mathrm{s}$