A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.00 m above the ground. The string breaks and the ball lands 2.30 m (horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the centripetal acceleration in m/s2 of the ball during its circular motion.

86.7 m/s²

Explanation:

Given that

Radius of the circle, r = 0.3 m

Height of the plane above the ground, = 1 m

Distance at which the ball lands, d = 2.3 m

To solve this question, we would use one of the principles of projectile motion.

v = ut + 1/2at²

-1 = 0 + 1/2 * -9.8 * t²

-1 = -4.9 t²

t² = 1/4.9

t² = 0.204

t = √0.204

t = 0.45 s

Then again, it's constant horizontal speed is

v(x) = x/t

v(x) = 2.3 / 0.45

v(x) = 5.1 m/s

And finally, the centripetal acceleration is gotten using the formula

a = v²/r

a = 5.1² / 0.3

a = 26.01 / 0.3

a = 86.7 m/s²