A heavy particle, suspended by a 1.5 m long string, is imparted a...

Let the string become slack when it makes an angle $\theta$ with the upward vertical.

At this instant, the tension in the string is zero. The component of gravity along the string provides the necessary centripetal acceleration:

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

$v^2 = gl \cos \theta$

Applying 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 $h$ reached by the particle from the lowest point is $h = l + l \cos \theta = l(1 + \cos \theta)$.

Substituting $h$ and $v^2$ into the energy equation:

$u^2 = gl \cos \theta + 2gl(1 + \cos \theta)$

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

Given $u = \sqrt{57} \text{ m/s}$, $l = 1.5 \text{ m}$, and $g = 10 \text{ m/s}^2$, we have $gl = 15 \text{ m}^2\text{/s}^2$. Substituting these values:

$57 = 2(15) + 3(15) \cos \theta$

$57 = 30 + 45 \cos \theta$

$45 \cos \theta = 27 \Rightarrow \cos \theta = \dfrac{3}{5}$

The speed of the particle at this instant is:

$v = \sqrt{gl \cos \theta} = \sqrt{15 \times \dfrac{3}{5}} = \sqrt{9} = 3.0 \text{ m/s}$