A boy whirls a stone in a horizontal circle 2 m above the ground by means of a string 1.25 m long. The string breaks and the stone flies off horizontally, striking the ground 10 m away. What is the magnitude of the centripetal acceleration ( in m/s²) during circular motion? Take g = 10 ms^-2

Given, h = 2 m, R = 1.25 m and horizontal distances = 10 m. When the string breaks, the stone is projected in the horizontal direction, which means that there is no initial vertical velocity.
From $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$, we have $(\because \mathrm{u}=0)$,
$\mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2 \left(\mathrm{i}\right)$
The horizontal distance travelled in time t is 
$\mathrm{s}=\mathrm{vt} \left(\mathrm{ii}\right)$
where v is the velocity of the stone in the horizontal direction which is the same as its velocity in circular motion.
Eliminating t from (i) and (ii) we get ${\mathrm{v}}^2=\frac{{\mathrm{gs}}^2}{2\mathrm{h}}$
Now, centripetal acceleration is 
${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}^2}{\mathrm{R}}=\frac{{\mathrm{gs}}^2}{2\mathrm{hR}}=\frac{10\times 100}{2\times 2\times 1.25}=200 {\mathrm{ms}}^{-2}$