Apendulum of length 0.350 m starts from rest at a maximum displacement of 10o from the equilibrium position…

a)what is the max speed of the pendulum?

b)at what time will the pendulum be located at an angle of displacement of 8o?

maximium speed     V = 0.32 m / s

Explanation:

The pendulum for small angles is a simple harmonic movement, in which the period is independent of the given angle and is described by the equation

    w = R l / g

a) To calculate the velocity write the energy at two points, higher and lowest  

   E1 = U = mg h

   E1= ½ m v²

How energy is conserved

   E1 = E2

   mgh = 1/2  m v²

   V = √2gh

Let's calculate the height by trigonometry

   cos θ = CA / H

   cos 10 = CA /0.35

   CA = 0.35 cos 10

   CA = 0.3447 m

The height that has risen Esla total height (0.35 m) minus this side

   H = L - CA

   H = 0.35 - 0.3447

   H = 0.0053 m

Let's calculate the speed

   V = √(2 9.8 0.0053)

   V = 0.32 m / s

Second part

It is requested to find the time for a certain angle, for this we use the equation of the simple harmonic movement

     Y (t) = A cos (wt)

Where A is the amplitude of the pendulum

    A = L - L cos T

    A = 0.35 (1- cos 10)

    A = 0.00532 m

    w = √g/L

    w = √ (9.8 / 0.35)

    w = 5.29 s⁻¹

We write the equation

   Y = 0.00532 cos (5.29 t)

To find time, let's use trigonometry to calculate the height of 8º

  Cos 8 = CA1 / L

   CA1 = L cos 8

   CA1 = 0.35 cos 8

   CA1 = 0.34659 m

the height raised

   Y = L -CA1

   Y = 0.35 -0.34659

   Y = 0.003406 m

We substitute in the equation of motion and clear time

  Y = 0.00532 cos (5.29 t)

  0.003406 = 0.00532 cos (5.29 t)

  cos (5.29 t) = 0.003406 / 0.00532 = 0.640

  5.29 t = cos⁻¹ 0.640

the angle in radians

   t =0.876 / 5.29

   t = 0.166 s

unable to answer the rest since your question is incomplete

and 1and 1