Question

bob of mass M is suspended by a massless string of length L. The horizontal velocity v at bottom most point is just sufficient to make it reach the top most point.. The angle θ with respect to vertical at which the speed is half that at initial point, satisfy

Answer 1

use conservation of energy ,
$\frac{1}{2}mv_0^2=\frac{1}{2}mv^2+mgl(1-cos\theta)$.....(1)
where $v_0$ is the horizontal velocity at the bottom point , v is the velocity of bob where bob inclined $\theta$ with vertical.

also we know, relation between velocity at the topmost/utmost point and velocity at the bottom point.
$mg(2l)=\frac{1}{2}mv_0^2-\frac{1}{2}mv_{top}^2$.........(2)

since, $v_0$ is just sufficient
$\frac{mv_{top}^2}{l}=T+mg$
T = 0, $v_{top}=\sqrt{gl}$

then equation (2),
$v_0=\sqrt{5gl}$

A/C to question, $v=\frac{v_0}{2}$

so, from equation (1),
$\frac{1}{2}m(5gl)=\frac{1}{2}m\left(\frac{5gl}{4}\right)+mgl(1-cos\theta)$

$\frac{20mgl-5mgl}{8}=mgl(1-cos\theta)$

$(1-cos\theta)=\frac{15}{8}$

$cos\theta=\frac{7}{8}$

hence, $\frac{3\pi}{4}<\theta<\pi$

Answer 2

So this is the answer

Hope this will HELP you..