Question

The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of √(3gl) . Find the angle rotated by the string before it comes slack.

Answer 1

Let, length of the string = l
bob speed = u =√mgl

{Refer to the attachment}

Now Total mechanical energy 

$1/2mv^2 -1/2mu^2 = -mgh$
$v^2 = u^2 -2g$ (l+ lcosθ)
$v^2 = 3gl - 2gl$ (1+cosθ) ---→ eq.(a)

Now consider the force here at the final position.
 mv² l = mg.cosθ
v² = lg.cosθ ---→ eq. (b)

Equating eq. (a) & (b) we get.

$3gl -2gl -$ 2gl cosθ =gl.cosθ .
By this we get,
3cos = 1
cosθ =1/3
___________
θ = cos⁻(1/3)
___________


At the time string become slack .
cos(180°-θ) = 1/3
___________
θ= cos⁻(-1/3)
___________


Hope it Helps :-)