Question

A body is rotated in the vertical plane by means of thread of length $4m$ with minimum possible speed. At the highest point B the thread breaks and the body moves under gravity and strikes at $C$. A is the lowest point of vertical circle which is almost touching ground. What is the horizontal range $AC$ (in $m$)?

Answer 1

Minimum possible speed is such that $\frac{{mv}^2}{r}=mg$
or $v=\sqrt{gl}=\sqrt{40}$ (because $R=l$)
This will be the velocity in horizontal direction at highest point (no vertical velocity at that point)
Now, range means when body reaches the bottom, i.e travels a distance of diameter of circular motion, i.e, $8m$ downwards.
Thus, $y=-8=u_{y}t-\frac{1}{2}g{t}^2-8=0-5t^{2}t=2\sqrt{\frac{2}{5}}sec$
Hence, $x=u_{x}t=2\sqrt{\frac{2}{5}}\times \sqrt{40}=8m$