Suppose the simple pendulum shown below were released from an angle of θ = 74°, with L = 0.73 m and m = 0.25 kg..What angle of release would give half the speed of that for the 74° release angle at the bottom of the swing?

Answer 1

Suppose the simple pendulum shown below were released from an angle of θ = 74°, with L = 0.73 m and m = 0.25 kg..What angle of release would give half the speed of that for the 74° release angle at the bottom of the swing?

Answer 2

You must have been given a formula to calculate this problem. What formulas do you have? You can't just expect to receive an answer without you asking what you need help on and jotting down the formulas, that's being pure lazy.

Answer 3

For half the speed at the bottom, you would need 1/4 the initial potential energy.

The initial P.E. is proportional to
(1 - cos A)
where A is the initial angle from vertical.
1 - cos74 = 0.7244
For 1/4 of that value,
1 - cosA = 0.1811
cosA = 0.8189
A = 35.0 degrees