Length of pendulum h = 1.5 m Let m be mass of the bob, then potential energy of the bob at A, $mgh = m \times 9.8 \times 1.5\,\mathrm{J}$On reaching the lowest point B, the bob will acquire an equal amount ofkinetic energy. But as 5% of energy is lost against the air friction.Energy converted $= 95\% (mgh)$If v is the velocity acquired by the bob at the point B, then$\text{K.E.} = \frac{1}{2} m v^{2} = \frac{95}{100} mgh$$v = \sqrt{ \frac{95}{100} \times 2 \times g h }$ $= \sqrt{ \frac{19}{20} \times 2 \times 9.8 \times 1.5 }$ $= 5.285\,\mathrm{m}\,\mathrm{s}^{-1}$