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

Here velocity at the lowest point `mu=sqrt(3gL)`
`sqrt(2gL)ltultsqrt(5gL`, the tension in the sting will be zero before velocity and it will leave circular path between `B` and `C` and then it will move as a free particle on a parabolic path.
`v^(2)=u^(2)-2gh`
`v^(2)=u^(2)-2gL(1+costheta)` .....(i)
`T+mg cos theta=(mv^(2))/(L)` .....(ii)
When the string becomes slack
`T=0` ....(iii)
Using (i),(ii)and (iii), we have
`gLcostheta=u^(2)-2gL(1+costheta)`
`=3gL-2gL(1+costheta)`
`cos theta=3-2-2cos theta`
`cos theta=(1)/(3)impliestheta=cos^(-1)((1)/(3))`
After `D`, the particle moves on a parabolic path
`H_(max)=(v^(2)sin^(2)theta)/(2g)=(v^(2)(1-cos^(2)theta))/(2g)`
`=((gL)/(3)(1-1/9))/(2g)=(L)/(6)xx(8)/(9)=(4L)/(27)`
maximum height attained by the bob above the point of suspension
`O=L cos theta+H_(max)=(L)/(3)+(4L)/(27)=(13L)/(27)`