Answer:
[tex]a_{cp}=162.2m/s^2[/tex]
Explanation:
The time the stone takes to fall can be calculated considering only the vertical component with the formula:
[tex]y=y_0+v_{0y}t+\frac{a_yt^2}{2}[/tex]
Taking the inital height as 0m and downward direction positive, since it departs from (vertical) rest we have:
[tex]y=\frac{gt^2}{2}[/tex]
Which gives us a time:
[tex]t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(2m)}{(9.8m/s^2)}}=0.64s[/tex]
Horizontally, on that time the stone travelled a distance x=10m, which means its horizontal speed was:
[tex]v_x=\frac{x}{t}=\frac{10m}{0.64s}=15.6m/s[/tex]
Since this speed is the tangential velocity while whirling, the centripetal acceleration of the stone was:
[tex]a_{cp}=\frac{v_x^2}{r}=\frac{(15.6m/s)^2}{(1.5m)}=162.2m/s^2[/tex]
