Kinematics/uniform circular motion.

[Muneerah]

Homework Statement

A ball on the end of a string is whirled around
in a horizontal circle of radius 0.28 m. The
plane of the circle is 1.44 m above the ground.
The string breaks and the ball lands 2.66 m
away from the point on the ground directly
beneath the ball’s location when the string
breaks.
The acceleration of gravity is 9.8 m/s² .
Find the centripetal acceleration of the ball
during its circular motion.
Answer in units of m/s².

Homework Equations

T= $$\sqrt{}2y/g$$
a_c=4*pi²r/T²

The Attempt at a Solution

First I found time using this equation T= $$\sqrt{}2y/g$$
=T= $$\sqrt{}2*1.44m/9.8m/s$$= .5421047s
then I plugged my time into the acceleration equation a_c=4*pi²r/T²
= 37.61415924 m. My answer is wrong and I would like to know what is it that I did wrong, I tried many ways to do it, but non of the answers I get are right, if you can please calrify. Thank you.

 

[kuruman]

The symbol T in the expression for the centripetal acceleration is not meant to be the time it takes the ball to hit the ground. It is the time the ball requires to complete one revolution, i.e. the period. T is what it is regardless of whether the string breaks or not. It is related to the speed v at which the ball goes around before the string breaks.

After the string breaks, you have projectile motion. Answer this question first, "How fast must the ball be moving in the horizontal direction so that it hits the ground 2.66 m away in 0.542 s?" This will give you the speed of the ball v. Knowing v, you can then find the centripetal acceleration. There is an expression that relates the centripetal acceleration to the speed. What is it?

 

[Muneerah]

Ok so basically to find the Velocity I used X/t = V
V= 4.907749077 m/s
centripetal acceleration = V^2/r
= 86.021432 m/s^2 and that is the right answer, thank you so much for helping me.