A rock is tossed straight up with a velocity of 33.9 m/s. When it returns, it falls into a hole 11.9 m deep. What is the rock's velocity as it hits the bottom of the hole?
To find the rock's velocity as it hits the bottom of the hole, we can use the laws of motion and apply the principle of conservation of energy.
First, let's analyze the motion of the rock. When the rock is thrown straight up, it experiences two main forces: gravity pulling it down and an initial velocity pushing it up. As the rock reaches its highest point, its velocity becomes zero. Then, gravity causes it to fall back down towards the ground.
We know that the initial velocity (v0) is 33.9 m/s, and the height of the hole (h) is 11.9 m. We want to find the final velocity (vf).
To solve for vf, we can use the principle of conservation of energy, which states that the total mechanical energy of a system remains constant.
At the highest point of the rock's motion, it has only potential energy, given by the equation:
Potential Energy (PE) = m * g * h
Where m is the mass of the rock, g is the acceleration due to gravity (which is approximately 9.8 m/s^2 near the Earth's surface), and h is the height of the hole.
Next, as the rock falls into the hole, it loses potential energy and gains kinetic energy. At the bottom of the hole, the energy is all kinetic, given by the equation:
Kinetic Energy (KE) = (1/2) * m * vf^2
Since the total mechanical energy is conserved (ignoring air resistance), the potential energy at the highest point is equal to the kinetic energy at the bottom point:
PE = KE
m * g * h = (1/2) * m * vf^2
We can cancel out the mass (m) on both sides of the equation:
g * h = (1/2) * vf^2
Now, we can rearrange the equation to solve for vf:
vf^2 = 2 * g * h
vf = sqrt(2 * g * h)
Plugging in the values, we get:
vf = sqrt(2 * 9.8 m/s^2 * 11.9 m)
vf = sqrt(230.8 m^2/s^2)
vf ≈ 15.2 m/s
Therefore, the rock's velocity as it hits the bottom of the hole is approximately 15.2 m/s.