Question

A 65.0 kilogram trampoline performer stands on a platform 3.00m above the trampoline. If the performer jumps with an initial vertical velocity of 5.00 m/s, find his speed as he impacts the trampoline. If the trampoline behaves like a spring with a spring constant of 6.2 x 10⁴ N/m, how far will the trampoline be compressed?

Answer 1

The speed of the trampoline performer as he impacts the trampoline is 7.0 m/s, and the compression distance of the trampoline is 0.106 m.

Step-by-step explanation:

To find the speed of the trampoline performer as he impacts the trampoline, we can use the principle of conservation of energy. The initial potential energy of the performer is determined by his height above the trampoline, while the final kinetic energy is determined by his speed. Since the trampoline behaves like a spring, the compression distance can be calculated using Hooke's Law.

Speed of the performer:

1. Initial potential energy: mgh = (65.0 kg)(9.8 m/s²)(3.00 m)
2. Final kinetic energy: (1/2)mv² = (1/2)(65.0 kg)v²

Setting the two equations equal to each other and solving for v, we get:

v = √(2gh) = √(2(9.8 m/s²)(3.00 m)) = 7.0 m/s

Compression distance of the trampoline:

Hooke's Law states that the force exerted by a spring is proportional to the amount it is compressed. We can use this to calculate the compression distance:

F = kx, where F is the force, k is the spring constant, and x is the compression distance.

Setting the gravitational force equal to the force exerted by the trampoline:

mg = kx

Solving for x:

x = mg/k = (65.0 kg)(9.8 m/s²)/(6.2 × 10⁴ N/m) = 0.106 m