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Question:

A stone is moved in a horizontal circle of radius 4m by means of a string at a height of 20m above the ground. The string breaks and the particle flies off horizontally, striking the ground 10m away. The centripetal acceleration during circular motion is

6.25ms⁻¹

18.75ms⁻¹

25ms⁻¹

12.5ms⁻¹

Solution:

The stone falls by height 20m. Hence the time of flight can be found out by
h=1/2at² ⇒t=√(2h/a)=√(2×20/10)s=2s
The range of the flight=vt=R ⇒v=10m/2s=5m/s
Hence the centripetal acceleration=v²/r=25m²/s²/4m=6.25m/s²