what is Bruce's speed at the bottom of the swing?

Question

Bruce stands on a bank beside a pond, grasps the end of a 10.0 m long rope attached to a nearby tree and swings out to drop into the water. If the rope starts at an angle of 38.0° with the vertical, what is Bruce's speed at the bottom of the swing?
____m/s

My biggest problem is I dont understand what 38.0degrees with the vertical means

Answer 1

Think of the point where the rope is tied to the tree branch. When Bruce is standing on the bank holding the rope, the rope makes an angle with a vertical line of 38 degrees measured at the point where the tree and rope meet. You can think of the rope as the hypotoneus of a right triangle. The triangle stands on one end so one leg is vertical. The angle between the hypotoneus and vertical leg is 38 deg.

Since the rope is the hypotoneus, and it is 10 m long. Bruce is:

h = 10*cos(38) = 8.2 m off the ground

He can descend a distance of d = 10 - 8.2 = 1.8 m before letting go of the rope. So by conservation of energy:

PE = mgd = KE = 1/2mv^2 we can find v.

v = sqrt(2gd) = sqrt(2*9.8*1.8) = 5.94 m/s