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A trapeze artist starts his motion with a trapeze at rest at an angle of \(45^\circ\) relative to the vertical. The trapeze rope has a length of 5 m. What is his speed at the lowest point of the trajectory?

Answer :

The speed of the trapeze artist at the lowest point of the trajectory is approximately 7.07 m/s.

To find the speed at the lowest point, we can use the principle of conservation of energy. At the highest point, all the potential energy is converted into kinetic energy.

Potential Energy (PE) at the highest point = Kinetic Energy (KE) at the lowest point

At the highest point, the only form of energy is potential energy given by:

PE = m * g * h

Where:

m = mass of the trapeze artist (assumed to be negligible)

g = acceleration due to gravity (approximately 9.8 m/s²)

h = height of the highest point (equal to the vertical displacement of the trapeze, which is 5 m)

At the lowest point, all the potential energy is converted into kinetic energy given by:

KE = (1/2) * m * v^2

Where:

v = speed of the trapeze artist at the lowest point (what we want to find)

Setting the potential energy at the highest point equal to the kinetic energy at the lowest point, we have:

m * g * h = (1/2) * m * v^2

Simplifying and solving for v, we get:

v = √(2 * g * h)

Plugging in the values, we get:

v = √(2 * 9.8 m/s² * 5 m) ≈ 7.07 m/s

Therefore, the speed of the trapeze artist at the lowest point of the trajectory is approximately 7.07 m/s.

to know more about principle of conservation of energy visit:

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