Fig 1 shows a pendulum of length L = 1.0 m. Its ball has speed of vo=2.0 m/s when the cord makes an angle of 30 degrees with the vertical. What
is the speed (V) of the ball when it passes the lowest position?

v = 2.57 m / s

Explanation:

For this exercise let's use conservation of energy

starting point. When it is at an angle of 30º

          Em₀ = K + U = ½ m v₁² + m g y₁

final point. Lowest position

          Em_f = K = ½ m v²

as there is no friction, the energy is conserved

          Em₀ = Em_f

          ½ m v₁² + m g y₁ = ½ m v²

Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle

         cos 30 = L ’/ L

         L ’= L cos 30

         y₁ = L -L '

          y₁ = L- L cos 30

we substitute

          ½ m v₁² + m g L (1- cos 30) = ½ m v²

           v =

let's calculate

           v =

           v = 2.57 m / s

newton's first law of motion

i don't know that i agree with that but what's the question ?