Bowling ball weighing 710 N attached to the ceiling by a rope of length 3.87 mThe ball is pulled to one side and released it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.51 m/s What is the acceleration of the bowling ball, in magnitude and direction, at this instant?

What is the tension in the rope at this instant?

a = 5.256 m / s²  with the vertical direction upwards and   T = 1090.8 N

Explanation:

to find the acceleration we must use Newton's second law, in the lowest part of the motion

          T - W = m a

where the acceleration is centripetal

         a = v² / r

let's calculate

         a = 4.51² / 3.87

         a = 5.256 m / s²

As the acceleration is centripetal it is directed towards the center of the circle, which in the lower part coincides with the vertical direction upwards.

Let's find the tension of the rope with the first equation

        T = W + m a

        W = m g

let's calculate

         T = 710 + 710 5.256 / 9.8

         T = 1090.8 N

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your answer is c! hope this : )