Question: A(n) 3.17 m long pendulum is released from

rest when the support string is at an angle of 38.6� degrees with the vertical.
The acceleration of gravity is 9.8 m/s2 .
What is the speed of the bob at the bottom
of the swing?
Answer in units of m/s

What I did:
y=3.17cos38.6 = 1.968297399
h=3.17m-1.968297399m =1.20170261m
v^2 =2(9.8)(1.20170261) =23.55337116
sqrt(23.55337116) =4.85318155

what did I do wrong and how do I fix it? thanks

y=3.17cos38.6 =3.17 •0.78=2.48

h= 3.17 - 2.48=0.69
v=3.68 m/s

Your approach is correct, but there is a small mistake in your calculation. Let's go through the steps again:

Step 1: Find the vertical displacement, y.
You correctly found y = 3.17 cos(38.6°) = 1.968297399 m.

Step 2: Find the height, h.
The height (h) is the difference between the total length of the pendulum (3.17 m) and the vertical displacement (y).
So, h = 3.17 m - 1.968297399 m = 1.201702601 m.

Step 3: Find the velocity at the bottom of the swing.
You correctly used the equation v^2 = 2gh, where g is the acceleration due to gravity (9.8 m/s^2) and h is the height.
So, v^2 = 2(9.8 m/s^2)(1.201702601 m) = 23.5533714 m^2/s^2.

Step 4: Take the square root to find the speed (magnitude of velocity).
You made a mistake in taking the square root. The correct square root of 23.5533714 is approximately 4.853181548 m/s (rounding to 5 significant digits).

Therefore, the correct speed of the bob at the bottom of the swing is approximately 4.853181548 m/s.