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 files off horizontally , striking the ground 10 m away . What is the magnitude of the centripetal accelerates during circular motion ? Take $g=10m{s}^{-2}$.a. $100m{s}^{-2}$ b. $200m{s}^{-2}$ c. $300m{s}^{-2}$ d. $400m{s}^{-2}$

Given, $h=2m$, $R=1.25m$ and horizontal distance $s=10m$. When the string breaks, the stone is projected in the horizontal direction, which means that there is no initial vertical velocity. From $s=ut+\frac{1}{2}g{t}^2$, we have $\left(\text{since }u=0\right)$$h=\frac{1}{2}g{t}^2$ ------(i) The horizontal distance travelled in time is $s=vt$ -----(ii) 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 $v^2=\frac{g{s}^2}{2h}$ Now, centripetal acceleration is $a_c=\frac{v^2}{R}=\frac{g{s}^2}{2hR}=\frac{10\times 100}{2\times 2\times 1.25}=200m/s^2$