A rock is tossed straight up with a velocity of 32.1 m/s. When it returns, it falls into a hole 20.1 m deep. What is the rock's velocity as it hits the bottom of the hole?

To determine the rock's velocity as it hits the bottom of the hole, we need to consider its motion when it returns from its highest point.

Let's analyze the problem using the kinematic equations of motion. The equation that relates an object's final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d) is as follows:

vf^2 = vi^2 + 2ad

In this case, when the rock is at its highest point, its final velocity is 0 m/s (since it changes its direction and starts to fall downward). The initial velocity is 32.1 m/s (given in the question). The acceleration due to gravity (a) is constant and is approximately -9.8 m/s^2 (negative because it acts downward). The displacement (d) is the depth of the hole, which is 20.1 m.

Now, let's substitute these values into the equation:

0^2 = (32.1)^2 + 2(-9.8)(20.1)

Simplifying the equation:

0 = 1030.41 - 392.04

392.04 = 1030.41

Now, subtract 392.04 from both sides:

vf^2 = 1030.41 - 392.04

vf^2 = 638.37

Finally, take the square root of both sides to find the final velocity, vf:

vf = √638.37

vf ≈ 25.27 m/s

Therefore, the rock's velocity as it hits the bottom of the hole is approximately 25.27 m/s.

The rock gains kinetic energy equal to the potential energy lost falling down the hole.

(M/2)[(Vfinal)^2 - 32.1)^2] = M*g*(depth)

Solve for Vfinal

The answer would be the same whether the rock was thrown up or down, initially.