Question

A circus performer swings down from a platform on a rope tied to the top of a tent in a pendulum-like swing. The performer's feet tough the ground 9 meters below where the rope is tied. how fast is the performer moving at the bottom of the arc?

Answer 1

The circus performer's speed at the bottom of the pendulum-like swing is approximately 13.28 m/s, calculated using the conservation of energy between potential and kinetic energy.

Step-by-step explanation:

The student is asking about the speed of a circus performer at the bottom of a pendulum-like swing. To solve this, we can apply the principles of conservation of energy, where potential energy at the top of the swing is converted into kinetic energy at the bottom. The potential energy (PE) when the performer is 9 meters above the ground is given by PE = mgh, where m is the mass of the performer, g is the acceleration due to gravity (9.8 m/s2), and h is the height (9 meters).

At the bottom of the swing, all the potential energy is converted into kinetic energy (KE), where KE = 1/2 m v2, and v is the velocity we want to find. Setting these equal to each other and solving for v gives:

mg h = 1/2 m v2
v = √(2gh)
v = √(2 × 9.8 m/s2 × 9 m)
v = √(176.4 m2/s2)
v = 13.28 m/s

Therefore, the circus performer's speed at the bottom of the arc is approximately 13.28 meters per second.

Answer 2

First solve for time to fall using: = √2 ∶ 2(9) = 1.35. Then solve for velocity √9.8

2

2

using: = ∶ 9.8 2 ∗ 1.35 =13.23 m/s The performer’s velocity is 13.2 m/s

at the bottom of the swing.

Step-by-step explanation: