5. An acrobat, starting from rest, swings freely on a trapeze of
length 3.7 m (Figure 6). If the initial angle of the trapeze is
48°, use the law of conservation of energy to determine
(a) the acrobat's speed at the bottom of the swing
(b) the maximum height, relative to the initial position, to
which the acrobat can rise

The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.

Em = K + U

Let's write the energy in two points.

Starting point. Highest part of the oscillation

Em₀ = U = m g h

Final point. Lower part of the movement

[tex]Em_f[/tex] = K = ½ m v²

Energy is conserved.

Emo = [tex]Em_f[/tex]

m g h = ½ m v²

v² = 2 gh

Let's use trigonometry to find the height, see attached.

h = L - L cos θ

h = L (1- cos θ)

They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.

h = 3.7 (1- cos 48)

h = 1.22 m

this is the maximum height of the movement.

Let's calculate the velocity.

[tex]v= \sqrt{2 \ 9.8 \ 1.22}[/tex]

v = 4.89 m / s

In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;

a) The maximum speed is v = 4.89 m / s

b) The maximum height is h = 1.22 m

Learn more here: brainly.com/question/13010190