Centripetal acceleration and tension

A child whirls a stone with a weight of 14 N in a horizontal circle 2.0 m above the ground by means
of a string 1.25 m long. The string breaks and the stone flies off horizontally, striking the ground 11m away.
(a) What was the centripetal acceleration of the stone while it was in circular motion?
(b) What was the breaking tension of the string?

Solve by considering horizontal and vertical motion of the free stone separately.

First calculate the time taken to fall to the ground when the string breaks. This is one of the standard kinematics equations.

s = ut + ½at²

where
s = distance to fall = 2.0m
u = initial velocity in a vertical direction = 0 m/s
a = acceleration = "g" = 9.81 m/s²
t = time

2.0 = 0 + ½ * 9.81 *t²
t² = 0.41 s²
t = 0.64 s

Horizontally, ignoring air resistance, the stone will travel at the same speed though out its flight.
It travels 11 m in 0.64 s. That's a speed of (11/0.64) = 17.2 m/s

So that was its tangential velocity v when the string broke.

(a)
Centripetal acceleration = v² / r
= 17.2² / 1.25
= 237 m/s²

(b)
Tension = centripetal force = mass * centripetal acceleration
mass = weight / g = 14 / 9.81 = 1.43 kg

Breaking tension = 1.43 * 237 = 339 N

Let X = Ux T so that Ux = X/T = 11/T is the tangential speed when the string breaks. We need to find T, the time the stone is in the air.

T = sqrt(2H/g) from H = 1/2 gT^2. So...

Ux = 11/T = 11/sqrt(2*2/9.8) = 17.21772343 = 17.2 mps and then Ar = Ux^2/R = 17.2^2/1.25 = 236.672 m/s^2 ANS a)

So we have Fr = mAr = (W/g)Ar = (14/9.8)*236.7 = 338.1428571 = 338.1 N ANS b)