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 s=ut+12gt2, we have (∵u=0),h=12gt2          (i)The horizontal distance travelled in time t 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 v2=gs22hNow, centripetal acceleration is ac=v2R=gs22hR=10×1002×2×1.25=200 ms−2