A grasshopper leaps from the ground with an initial speed of 2 m/s...

Given: initial speed $u = 2\text{ m/s}$, angle of projection $\theta = 60^{\circ}$, and $g = 10\text{ m/s}^2$.

The maximum height $H$ attained by a projectile is given by the formula:
$H = \dfrac{u^2 \sin^2 \theta}{2g}$

Substitute the given values into the formula:
$\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}$
$\sin^2 60^{\circ} = \dfrac{3}{4}$

$H = \dfrac{(2)^2 \times \left(\dfrac{3}{4}\right)}{2 \times 10}$
$H = \dfrac{4 \times \dfrac{3}{4}}{20}$
$H = \dfrac{3}{20}\text{ m}$

To convert the height into centimeters, multiply by $100$:
$H = \dfrac{3}{20} \times 100\text{ cm} = 15\text{ cm}$.

Answer: $15\text{ cm}$