A rock is tossed straight up with a velocity of 34.0 m/s. When it returns, it falls into a hole 17.3 m deep. What is the rock's velocity as it hits the bottom of the hole?

How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

At the bottom of the hole, kinetic energy is increased by M g D from the original value. D is the depth.

Therefore V2^2/2 - V1^2.2 = g D
(The mass cancels out).
Solve for V2, which will be negative (going down)
V1 is the initial upward velocity

For the time in the air, T, solve

V1*T - (g/2)T^2 = -17.3 meters

To find the rock's velocity as it hits the bottom of the hole, we can use the principle of conservation of energy.

First, let's analyze the initial and final energy states of the rock. At the bottom of the hole, the rock is at rest, so its gravitational potential energy is zero. At the highest point of its trajectory, the rock comes to rest momentarily, so its kinetic energy is zero.

Now, let's calculate the initial and final kinetic energies of the rock. At the bottom of the hole, the kinetic energy is given by:

Kinetic Energy = (1/2) * mass * velocity^2

Since the rock is at rest, the kinetic energy is zero. Therefore, we have:

0 = (1/2) * mass * velocity^2

Simplifying, we find:

velocity^2 = 0

Taking the square root of both sides, we get:

velocity = 0 m/s

Therefore, the rock's velocity as it hits the bottom of the hole is 0 m/s.

Next, let's determine the time it takes for the rock to hit the bottom of the hole. We can use the equations of motion to find the time.

The relationship between time (t), initial velocity (u), final velocity (v), and acceleration (a) is given by:

v = u + at

In the case of the rock thrown straight up, its initial velocity (u) is 34.0 m/s, its final velocity (v) is 0 m/s (when it reaches the highest point), and the acceleration (a) is the acceleration due to gravity (-9.8 m/s^2, taking into account its direction).

0 = 34.0 m/s + (-9.8 m/s^2) * t

Solving for t, we get:

34.0 m/s = 9.8 m/s^2 * t

t = 34.0 m/s / 9.8 m/s^2

t ≈ 3.47 seconds

Therefore, the rock is in the air for approximately 3.47 seconds, from the instant it is released until it hits the bottom of the hole.