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 1.50 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 13.3 m/s.

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

To determine the gymnast's initial speed coming off the trampoline, we can use the principle of conservation of mechanical energy. The mechanical energy of the gymnast is conserved as long as there are no external forces acting on them.

The mechanical energy of the gymnast at the highest point of the jump (when they pass the window) consists of two components: potential energy (PE) and kinetic energy (KE). At this point, all of the initial potential energy is converted into kinetic energy.

The potential energy (PE) can be calculated using the equation:

PE = m * g * h

where m is the mass of the gymnast, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the window above the trampoline (1.50 m).

The kinetic energy (KE) can be calculated using the equation:

KE = 0.5 * m * v²

where v is the final velocity (13.3 m/s) of the gymnast as they pass the window.

Since the mechanical energy is conserved, the initial potential energy (PE) must be equal to the final kinetic energy (KE). Therefore, we can equate the two equations:

m * g * h = 0.5 * m * v²

Mass (m) cancels out on both sides of the equation. We can rearrange the equation to solve for the initial velocity (u):

u = sqrt((2 * g * h))

Substituting the given values:

u = sqrt((2 * 9.8 * 1.50))

Simplifying the equation:

u = sqrt(29.4)

Calculating the square root:

u ≈ 5.42 m/s

Therefore, to reach a speed of 13.3 m/s as the gymnast passes the window on their way down, their initial speed coming off the trampoline must have been approximately 5.42 m/s.

To find the gymnast's initial speed coming off the trampoline, we can make use of the principle of conservation of mechanical energy. Since the gymnast went up past the window and then came back down, we can assume that at the highest point of his jump, all his initial potential energy is converted into kinetic energy.

At the highest point, the gymnast's potential energy is equal to his gravitational potential energy, which can be calculated using the formula:

PE = m * g * h

where m is his mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the trampoline (1.50 m in this case).

At the highest point, all the potential energy is converted to kinetic energy, which can be calculated using the formula:

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

where v is his speed at this point (13.3 m/s).

Since the energy is conserved, we can equate the two expressions for energy:

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

Simplifying this equation, we find:

2 * g * h = v^2

Plugging in the given values:

2 * 9.8 m/s^2 * 1.50 m = v^2

29.4 m^2/s^2 = v^2

Taking the square root of both sides, we find:

v ≈ √29.4 m/s

So, the gymnast's initial speed coming off the trampoline was approximately 5.42 (rounded to two decimal places) m/s.