The bob of a simple pendulum is given a sharp hit impart it a horizontal speed of `sqrt(3gL)`. Find an angle made by the string with the upper verticle before it becomes slack. Also, calculate the maximum height attained by the bob above the point of suspension. `L` is length of the string.

Explanation

Now, after slackening of the string, the motion of the bob is under gravity only, for which the maximum height from P is given by
`h_(2)=(v^(2)sin^(2)alpha)/(2g)`
where `v^(2)=(gl)/(3)andsin^(2)alpha=1-cos^(2)alpha=1-((1)/(3))^(2)=(8)/(9)`
`therefore h_(2)=(((gl)/(3))((8)/(9)))/(2g)=(4l)/(27)`
`therefore` maximum height from A is = `h_(1)+h_(2)`
`=l(1+(1)/(3))+(4l)/(27)=(40l)/(27)`