Suspended by a 1.5 m long string, a heavy particle is given a...

Let $\theta$ be the angle made by the string with the upward vertical when it becomes slack.

By conservation of mechanical energy between the lowest point and the point where the string becomes slack:

$\dfrac{1}{2}mu^2 = \dfrac{1}{2}mv^2 + mgh$

The height $h$ of the particle from the lowest point is $l + l\cos\theta$.

$v^2 = u^2 - 2g(l + l\cos\theta)$

At the point where the string becomes slack, the tension $T = 0$. The component of weight along the string provides the necessary centripetal force:

$mg\cos\theta = \dfrac{mv^2}{l}$

$v^2 = gl\cos\theta$

Substituting $v^2$ into the energy equation:

$gl\cos\theta = u^2 - 2gl - 2gl\cos\theta$

$3gl\cos\theta = u^2 - 2gl$

$\cos\theta = \dfrac{u^2 - 2gl}{3gl}$

Given $u = \sqrt{57} \text{ m/s}$, $l = 1.5 \text{ m}$, and $g = 10 \text{ m/s}^2$:

$\cos\theta = \dfrac{57 - 2(10)(1.5)}{3(10)(1.5)}$

$\cos\theta = \dfrac{57 - 30}{45} = \dfrac{27}{45} = \dfrac{3}{5}$

\theta = 53^{}^{{}^{\circ}}