A person on a trampoline bounces straight upward with an initial speed of 4.7m/s . What is the magnitude of the person's speed when she returns to her initial height?

She will return to the trampoline at the same speed. Conservation of energy tells you that. (Air resistance is neglected)

To find the magnitude of the person's speed when she returns to her initial height, we can use the principle of conservation of mechanical energy.

The mechanical energy of the person-trampoline system at the initial height and at the maximum height (when the person reaches the highest point during the bounce) remains constant, assuming no energy losses due to factors like air resistance or friction.

At the initial height, the only form of energy is the potential energy (PE) due to the person's height from the ground. At the maximum height, the only form of energy is the kinetic energy (KE) due to the person's velocity.

Using the conservation of mechanical energy, we have:

Initial potential energy (PEi) = Maximum kinetic energy (KEf)

PEi = m * g * h, where m is the mass of the person, g is the acceleration due to gravity, and h is the initial height from the ground.

KEf = 0.5 * m * v^2, where v is the magnitude of the person's velocity at the maximum height.

Since the person returns to the initial height, the maximum height reached is equal to the initial height (h). Therefore:

PEi = KEf

m * g * h = 0.5 * m * v^2

By canceling out the mass (m) on both sides, the equation becomes:

g * h = 0.5 * v^2

Rearranging the equation to solve for v, we have:

v^2 = 2 * g * h

Taking the square root of both sides gives us:

v = sqrt(2 * g * h)

Now we can plug in the values to calculate the magnitude of the person's speed when she returns to her initial height. The acceleration due to gravity, g, is approximately 9.8 m/s^2, and the height, h, is presumably the height of the trampoline when the person initially jumped.