Question

A simple pendulum consists of a small ball attached to the end of a string of length 2.00m. The Pendulum is displaced to an angle of 60.0 degrees from the vertical and given an initial speed of 4.00m/s. What is the speed of the ball (in m/s) when it passes through the lowest point in its path?

Answer 1

Make a right triangle with the pendulum ball and the y-axis line. The hypotenuse is 2m long and using cos60 = adjacent / hyp,
0.5 = adj / 2
adjacent side = 1m
Since you know the ball at the bottom of the arc will be 2m long, then that means the ball will fall 1 meter more in vertical fall from the release point.

Vertical fall = 1/2 g * t^2
solving for t = time = 0.45 seconds to fall 1m to plumb bottom point

Velocity = gt
Velocity = 9.8 m/s^2 * 0.45s = 4.43 m/s

The velocity of 4.43 m/s is the velocity if you dropped a mass in freefall for one meter. Essentially when you release the ball at the 60 degree angle mark the x-axis crosses the y-axis at 1m, so the ball drops another meter to reach the bottom of the arc, and the only force acting on the ball is gravity.

Now just turn the problem to velocity vectors:

The ball has initial velocity of 4 m/sand direction is a tangent line from the ball end to 60 degrees downward from the y-axis. Gravity causes another downward vector with velocity of 4.43 m/s.
Put the vectors end to end and the resultant vector is 180 degree - 60 degree = 120 degree south of the x-axis.

Vector A = -120 degrees at 4 m/s and vector B = -90 degree at 4.43 m/s

Resultant = 8.1 m/s velocity at bottom of arc

Answer 2

I tried your method but unfortunately the result does not match the correct answer to this problem. Apparently the answer was 5.97 m/s. I don't understand how my instructor got that answer though. Heres a visual of the problem im trying to get.