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

Given:
Initial velocity, $u = \sqrt{57} \text{ m/s}$
Length of the string, $l = 1.5 \text{ m}$
Acceleration due to gravity, $g = 10 \text{ m/s}^2$

The minimum velocity required to complete the vertical circle is $\sqrt{5gl} = \sqrt{5 \times 10 \times 1.5} = \sqrt{75} \text{ m/s}$.
The minimum velocity required to reach the horizontal position is $\sqrt{2gl} = \sqrt{2 \times 10 \times 1.5} = \sqrt{30} \text{ m/s}$.

Since $\sqrt{30} < \sqrt{57} < \sqrt{75}$, the particle will cross the horizontal position but the string will become slack before reaching the highest point.

Let the string become slack at an angle $\theta$ with the upward vertical. At this point, the tension in the string is zero ($T = 0$).
The component of gravity provides the necessary centripetal force:
$mg \cos\theta = \dfrac{mv^2}{l} \Rightarrow v^2 = gl \cos\theta$

Using conservation of mechanical energy from the lowest point to the point where the string becomes slack:
$\dfrac{1}{2}mu^2 = \dfrac{1}{2}mv^2 + mgh$
Here, the height from the lowest point is $h = l(1 + \cos\theta)$.
$u^2 = v^2 + 2gl(1 + \cos\theta)$

Substituting $v^2 = gl \cos\theta$ and the given values:
$57 = 15 \cos\theta + 2(15)(1 + \cos\theta)$
$57 = 15 \cos\theta + 30 + 30 \cos\theta$
$27 = 45 \cos\theta \Rightarrow \cos\theta = \dfrac{27}{45} = \dfrac{3}{5}$

The velocity at this point is:
$v = \sqrt{15 \times \dfrac{3}{5}} = \sqrt{9} = 3 \text{ m/s}$

The height of this point above the point of suspension is:
$y = l \cos\theta = 1.5 \times \dfrac{3}{5} = 0.9 \text{ m}$

After the string becomes slack, the particle undergoes projectile motion. The velocity vector makes an angle $\theta$ with the horizontal. The vertical component of velocity is:
$v_y = v \sin\theta$
Since $\cos\theta = \dfrac{3}{5}$, we have $\sin\theta = \dfrac{4}{5}$.
$v_y = 3 \times \dfrac{4}{5} = 2.4 \text{ m/s}$

The further maximum height reached during the projectile motion is:
$H' = \dfrac{v_y^2}{2g} = \dfrac{(2.4)^2}{2 \times 10} = \dfrac{5.76}{20} = 0.288 \text{ m}$

The total maximum height reached by the particle above the point of suspension is:
H_{\text\max} = y + H' = 0.9 + 0.288 = 1.188 \text{ m}

Rounding to one decimal place, we get H_{\text\max} \approx 1.2 \text{ m}.

Answer: $1.2 \text{ m}$