$$KE = \frac{1}{2}mv^2$$
Where KE is kinetic energy, m is the mass of the rock, and v is its velocity.
First, we need to find the velocity of the rock at the midway point. We can use the equation of motion:
$$v^2 = u^2 + 2as$$
Where:
- v is the final velocity (at the midway point)
- u is the initial velocity (0 m/s, since the rock is dropped)
- a is the acceleration due to gravity (-9.8 m/s²)
- s is the distance traveled (half of the total height, 25 meters)
Plugging in the values, we get:
$$v^2 = 0 + 2(-9.8)(25)$$
$$v^2 = -490$$
$$v = \sqrt{-490} = 22.14 \ m/s$$
Now we can calculate the kinetic energy at the midway point:
$$KE = \frac{1}{2}(98)(22.14)^2$$
$$KE = 24,100 \ J$$
Therefore, the kinetic energy of the rack at the midpoint of its fall is 24,100 Joules.
