Answer :
The acrobat's speed at the bottom of the swing is 8.52 m/s.
To calculate the acrobat's speed at the bottom of the swing, we can use the principle of conservation of mechanical energy. At the highest point of the swing, all of the potential energy is converted into kinetic energy. At the bottom of the swing, all of the potential energy is converted back into kinetic energy.
The potential energy at the highest point of the swing can be calculated using the equation:
PE = m * g * h
Where m is the mass of the acrobat, g is the acceleration due to gravity (approximately 9.8 m/[tex]s^2[/tex]), and h is the height above the bottom of the swing.
In this case, the height above the bottom of the swing is equal to the length of the trapeze, since the acrobat starts from rest. So we can rewrite the equation as:
PE = m * g * L
Where L is the length of the trapeze.
At the highest point of the swing, all of the potential energy is converted into kinetic energy, which can be calculated using the equation:
KE = 1/2 * m * [tex]v^2[/tex]
Where v is the velocity at the bottom of the swing.
Since mechanical energy is conserved, we can equate the potential energy at the highest point of the swing to the kinetic energy at the bottom of the swing:
PE = KE
m * g * L = 1/2 * m * [tex]v^2[/tex]
Canceling out the mass on both sides of the equation gives us:
g * L = 1/2 * [tex]v^2[/tex]
Solving for v gives us:
v = √(2 * g * L)
Plugging in the values, with g = 9.8 m/[tex]s^2[/tex] and L = 3.7 m, we get:
v = √(2 * 9.8 * 3.7)
v = √(72.52)
v ≈ 8.52 m/s
Therefore, the acrobat's speed at the bottom of the swing is approximately 8.52 m/s.
To know more about speed refer here:
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