You are thinking about taking gymnastics, so you go to the facility and get an idea of what to expect by looking out from the viewing room. The viewing room window is 4.40 m above the trampoline directly below, so it is perfect for viewing the the facility. Occasionally someone jumps past the window and then back down. On one occurrence a gymnast went up past the window and came back down; as he passed the window on the way down, you notice that his speed is 15.3 m/s.

Question

You are thinking about taking gymnastics, so you go to the facility and get an idea of what to expect by looking out from the viewing room. The viewing room window is 4.40 m above the trampoline directly below, so it is perfect for viewing the the facility. Occasionally someone jumps past the window and then back down. On one occurrence a gymnast went up past the window and came back down; as he passed the window on the way down, you notice that his speed is 15.3 m/s.

What must have been his initial speed coming off the trampoline?

Answer 1

See previous post: Thu, 9-15-16, 10:47 PM.

Answer 2

To find the initial speed of the gymnast coming off the trampoline, we can use the principle of conservation of mechanical energy. The total mechanical energy of the gymnast-trampoline system remains constant, neglecting any losses due to air resistance or friction.

Since the gymnast started at the trampoline and reached a maximum height at the window level, we can consider the initial and final positions as the trampoline and the window. The initial kinetic energy of the gymnast is equal to the final potential energy at the window level.

Let's denote the initial speed of the gymnast as v0, the final speed at the window level as vf (15.3 m/s), the height of the window above the trampoline as h (4.40 m), and the acceleration due to gravity as g (approximately 9.8 m/s²).

At the trampoline, the kinetic energy of the gymnast is (½)mv0², where m is the mass of the gymnast.

At the window level, the potential energy of the gymnast is mgh, where h is the height of the window above the trampoline.

According to the conservation of mechanical energy, the initial kinetic energy is equal to the final potential energy:

(½)mv0² = mgh

We can cancel out the mass (m) from both sides of the equation:

(½)v0² = gh

Now, solve for the initial speed (v0):

v0² = 2gh

v0 = √(2gh)

Substituting the given values:

v0 = √(2 * 9.8 m/s² * 4.40 m)

v0 ≈ √(86.24) ≈ 9.30 m/s

Therefore, the initial speed of the gymnast coming off the trampoline must have been approximately 9.30 m/s.