91Ó°ÊÓ

The radius of the circular path is r = 1.5 m

The height at which stone whirls is y = 2.0 m

The horizontal distance traveled by stone after the breaking of the string is x = 10m

When the stone flies off horizontally and strikes the ground, it gets projectile motion. The stone falls from the maximum height y to the ground after covering a horizontal distance as x. At maximum height, the ball has zero vertical velocity; it has only horizontal velocity in the outward direction. By using kinematical equations, find horizontal velocity. We can use the expression of the centripetal acceleration in terms of velocity and radius.

Formula:

$$\begin{gathered} y=v_{0y}t+\frac{1}{2}g{t}^2 \\ x=v_{0t}t \\ a=\frac{v^2}{r} \end{gathered}$$

The stone is moving in a horizontal circle. When the string breaks, it flies off as shown inthefigure. It falls from maximum height, so vertical velocity$v_{0y}$of stone is zero, and it hasonly horizontal velocity. We can usethesecond kinematical equation along the vertical direction and find the time taken by the stone as,

$-y=v_{oy}t-\frac{1}{2}g{t}^2$

We can use the sign convention according to the motion of stone,

$t^2=\frac{2y}{g}$ …(¾±)

The horizontal distance covered by the stone is,

$$\begin{gathered} x=v_{x}t \\ {v}_x=\frac{x}{t} \end{gathered}$$

…(¾±¾±)

Using equation (i) and (ii), the stone is moving in horizontal circle with velocity$v_x=v$then the centripetal acceleration is,

$$\begin{gathered} a=\frac{v^2}{r} \\ =\frac{x^2}{t^{2}r} \\ =\frac{g{x}^2}{2yr} \\ =\frac{(9.8\mathrm{m}/{\mathrm{s}}^2)×{\left(10\mathrm{m}\right)}^2}{2×\left(2\mathrm{m}\right)×(1.5\mathrm{m})} \\ =163\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The magnitude of centripetal acceleration of the stone is $163m/s^2≈160m/s^2$.