A 55 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.5 m/s. (Ignore small changes in gravitational potential energy.)

(a) How fast is he going as he lands on the trampoline, 3.0 m below?
(b) If the trampoline behaves like a spring of spring constant 5.2 104 N/m, how far does he depress it?

find total height

h=(1/2)v^2*g
plug into equation v=sqrt(2*9.8*total height) that's velocity

To solve this problem, we can use the principle of conservation of mechanical energy. The initial mechanical energy of the trampoline artist is equal to the final mechanical energy when he lands on the trampoline. The mechanical energy includes kinetic energy and potential energy.

(a) First, let's calculate the final velocity of the trampoline artist as he lands on the trampoline, 3.0 m below:

Initial kinetic energy = 1/2 * mass * initial velocity^2
Final potential energy = mass * g * height (gravitational potential energy)

Since the artist jumps vertically upward, the height above the trampoline is given by:
Height above trampoline = Initial potential energy / (mass * g)

Final potential energy = mass * g * height below the trampoline
Final gravitational potential energy = mass * g * height below the trampoline

Equating the initial kinetic energy to the sum of final potential energy and final kinetic energy, we have:
1/2 * mass * initial velocity^2 = mass * g * height above trampoline + mass * g * height below trampoline + 1/2 * mass * final velocity^2

We can simplify this equation by canceling out the mass term:

1/2 * initial velocity^2 = g * height above trampoline + g * height below trampoline + 1/2 * final velocity^2

Rearranging the equation to solve for the final velocity:
1/2 * final velocity^2 = 1/2 * initial velocity^2 - g * (height above trampoline + height below trampoline)

Substituting the values:
Initial velocity = 5.5 m/s
Height above trampoline = 3.0 m
Height below trampoline = 3.0 m
g = 9.8 m/s^2

1/2 * final velocity^2 = 1/2 * (5.5 m/s)^2 - 9.8 m/s^2 * (3.0 m + 3.0 m)

1/2 * final velocity^2 = 15.125 m^2/s^2 - 58.8 m^2/s^2

1/2 * final velocity^2 = -43.675 m^2/s^2

Final velocity^2 = -87.35 m^2/s^2

Taking the square root of both sides:
Final velocity = √(-87.35 m^2/s^2) (Note: Since the velocity is negative, it means the artist is moving downward)

Therefore, the final velocity as he lands on the trampoline is approximately -9.35 m/s.

(b) Next, let's calculate how far he depresses the trampoline.

The trampoline behaves like a spring with a spring constant of 5.2 * 10^4 N/m. The potential energy stored in a spring is given by:

Potential energy of the spring = 1/2 * k * x^2

Where k is the spring constant, and x is the displacement from the equilibrium position.

The final potential energy in the trampoline is equal to the potential energy stored in the spring.

Initial potential energy = Final potential energy
mass * g * height below the trampoline = 1/2 * k * x^2

We can solve for x by rearranging the equation:

x^2 = (2 * mass * g * height below the trampoline) / k

Substituting the values:
mass = 55 kg
g = 9.8 m/s^2
height below trampoline = 3.0 m
k = 5.2 * 10^4 N/m

x^2 = (2 * 55 kg * 9.8 m/s^2 * 3.0 m) / (5.2 * 10^4 N/m)

x^2 = 0.060 m^2

Taking the square root of both sides:
x = √(0.060 m^2)

Therefore, the trampoline depresses by approximately 0.245 m when the artist lands on it.

To answer part (a) of the question, we can use the principle of conservation of energy. Here, we know the initial kinetic energy of the trampoline artist (K1) as he jumps from the platform and the potential energy (P1) at the top of the platform.

Step-by-step solution to part (a):
1. Determine the potential energy at the top of the platform (P1):
P1 = m * g * h1
where m = mass of the artist = 55 kg,
g = acceleration due to gravity = 9.8 m/s^2,
and h1 = height of the platform = 0 m (since the artist starts at the top).

2. Calculate the initial kinetic energy (K1):
K1 = (1/2) * m * v1^2
where v1 = initial velocity = 5.5 m/s.

3. Use the conservation of energy to find the total mechanical energy at the top (E1):
E1 = K1 + P1

4. Determine the kinetic energy (K2) of the artist as he lands on the trampoline.
Since the trampoline is 3.0 m below the starting point, the potential energy (P2) at that height is:
P2 = m * g * h2
where h2 = 3.0 m.

Therefore, K2 = E1 - P2

5. Calculate the final velocity (v2) of the artist as he lands on the trampoline.
To do this, we equate the kinetic energy to K2:
K2 = (1/2) * m * v2^2

Rearranging the equation:
v2^2 = (2 * K2) / m
v2 = √((2 * K2) / m)

Now, let's move to part (b) of the question. We need to calculate how far the trampoline depresses when the artist lands on it.

Step-by-step solution to part (b):
1. Use Hooke's Law to find the force exerted on the trampoline when the artist lands.
F = k * d
where k = spring constant = 5.2 * 10^4 N/m.

2. Calculate the force exerted on the trampoline (F):
Since the artist is standing on the trampoline, F = m * g.

3. Equate the force in step 1 and step 2:
F = m * g = k * d

Rearranging the equation:
d = (m * g) / k

Therefore, the depression distance (d) of the trampoline can be calculated using the formula above.

Note: In both parts (a) and (b), we are assuming there is no other external force acting on the trampoline artist.