A particle is suspended from a fixed point by a string of lengh `5m`. It is projected from the equilibrium position with such a velocity that the string slakens after the particle has reached a height `8m` above lowest point. Find the velocity of the particle, just before the string slackens.
Find also, to what to what height the particle can rise further?

Explanation

As shown in figure, at point B, `T=0`
Thus, we have
`(mv^2)/(l)=mg cos theta`
`v=sqrt(glcostheta)=sqrt(29.4)=5.42ms^-1`
After point B, particle will move in a projectile trajectory with this initial velocity. Let h be the maximum height it further rises. It is given as
`h=(v^2sin^2theta)/(2g)=(29.4xx(4//5)^2)/(2xx9.8)=0.96m`