Final answer:
Using kinematic equations, the trampoline artist is calculated to be moving at approximately 4.36 m/s as he lands on the trampoline.
Step-by-step explanation:
To determine how fast the trampoline artist is going as he lands on the trampoline, we can use the principles of conservation of energy or kinematic equations. Considering the question provides a vertical displacement and initial velocity, it's appropriate to use the kinematic equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and s is the displacement.
Given that the artist jumps off with an initial velocity (u) of 4.5 m/s upwards and falls 2.0 m downwards to the trampoline, we would take the acceleration due to gravity (a) as positive 9.81 m/s^2 (since the direction of gravity is the same as the direction of displacement), and the displacement (s) as -2.0 m. By substituting the values into the equation we get:
v^2 = u^2 + 2asv^2 = (4.5 m/s)^2 + 2(9.81 m/s^2)(-2.0 m)
We need to solve for v, which will give us the final velocity.
Calculate the initial kinetic energy: (4.5 m/s)^2 = 20.25 m^2/s^2
Calculate the potential energy change: 2(9.81 m/s^2)(-2.0 m) = -39.24 m^2/s^2
Add the kinetic and potential energy changes: 20.25 m^2/s^2 - 39.24 m^2/s^2 = -18.99 m^2/s^2
Since the square of the final velocity cannot be negative, the minus sign indicates a change in direction, so we take the positive root: v = √(18.99 m^2/s^2) ≈ 4.36 m/s
The trampoline artist would be going roughly 4.36 m/s as he lands on the trampoline.