High School

An aerialist on a high platform holds onto a trapeze attached to a support by a 9.1 m cord. Just before he jumps off the platform, the cord makes an angle of 39° with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.80 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.

A) 51.5°

B) 39°

C) 51.3°

D) 38.7°

Answer :

Final Answer:

51.3° is the angle θ that the trapeze cord makes with the vertical at this instant.

Therefore, the correct answer is C) 51.3°.

Explanation:

The angle θ can be determined using the conservation of mechanical energy. At the highest point of the swing, all the initial gravitational potential energy is converted into kinetic energy.

At the lowest point, this kinetic energy is entirely transformed back into gravitational potential energy. The initial height can be calculated using trigonometry:

[tex]\(h = 9.1 \times \sin(39°[/tex]

The potential energy at the lowest point is then given by:

[tex]\(PE = m \times g \times (9.1 - 0.8)\)[/tex]

Equating the potential energy at the highest and lowest points:

[tex]\(m \times g \times (9.1 - 0.8) = m \times g \times 9.1 \times \sin(39°)\)[/tex]

The mass 'm' cancels out, and we can solve for sin(39°). Taking the inverse sine of this value gives the angle θ:

[tex]\(θ = \sin^{-1}\left(\frac{(9.1 - 0.8)}{9.1}\right)\)[/tex]

Calculating this gives θ ≈ 51.3°.

Therefore, the correct answer is C) 51.3°.