The bob A of a simple pendulum is released from a horizontal position A as shownin in figure. If the length of the pendulum is 1.5m , what is the speed with which the bob arrives at the lowermost point B, given that it dissipates `5%` of its initial energy against air resistance ?

Explanation

On releasing the bob of pendulum from horizontal position, it falls vertically downward by a distance equal to length of pendulum i.e. H=I=1.5m
As `5%` of loss in P.E is disspiated against air resistance, the balance 95% energy is transformed into K.E. Hence,
`1/2 mv^(2) = 95/100 xx mgh`
`rArr v= sqrt(2 xx 95/100 xx gh) = sqrt(2 xx 95 xx 9.8 xx 1.5)/100 = 5.3 ms^(-1)`