A pendulum bob has a speed 3m/s while passing thorugh its lowest position. What is its speed when it makes an angle of `60^0` with the vertical? The length of the pendulum is 0.5m Take `g=10 m/s^2`.

Explanation

Refer to Figure, In moving from `A` to `B` , the bob rises through
`AC=OA-OC=l-l costheta=l(1-costheta)`
If `upsilon_(1)` is velocity at `B` , then applying principle of conservation of energy.
`(1)/(2)m upsilon_(1)^(2)=(1)/(2)m upsilon^(2)-mgl(1-costheta)`
`upsilon_(1)^(2)=upsilon^(2)-2gl(1-costheta)`
`(=3^(2)-2xx10xx0.5(1-cos60^(@))`
`=9-5=4`
`upsilon_(1)=sqrt(4)=2m//s`