Question

5. An acrobat, starting from rest, swings freely on a trapeze of

Answer 1

The energy conservation 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

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

            $Em_f$ = K = ½ m v²

Energy is conserved.

            Emo = $Em_f$  

            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.  

          $v= \sqrt{2 \ 9.8 \ 1.22}$  

          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