A simple pendilum with bob of mass m and length l is held in position at an angle `theta` with t he verticle. Find its speed when it passes the lowest position.

Explanation

The height bob comes down is QR
`h=QR=SQ=-SR=I-I costheta`
By law of conservation of mechanical energy
`mgh=1/2mv^(2)`

During motion of the bob from P to Q, its PE gets converted to KE. The height h, it comes in downward direction is QR
`h=QR=SQ-SR=I-Icos theta.`
By the law of conservation of mechanical energy
`mgh=1/2mv^(2)`
`mg(-lcostheta)=1/2mv^(2)`
`v=sqrt(2gl(1-costheta))`