Question

A Frisbee gets stuck in a tree. You want to get it out by throwing a 1.0-kg rock straight up at the Frisbee. If the rock’s speed as it reaches the Frisbee is 4.0 m/s, what was its speed as it left your hand 2.8 m below the Frisbee? Specify the system and the initial and final states.

Answer 1

The speed of the rock as it left your hand was 8.18 m/s.

Step-by-step explanation:

To find the speed of the rock as it left your hand, we can use the principle of conservation of mechanical energy. The system can be defined as the rock and the Earth, with the initial state being when the rock is in your hand and the final state being when the rock reaches the Frisbee in the tree.

Since the rock is thrown straight up, its initial velocity is the same as its final velocity when it reaches the Frisbee. We can use the equation for gravitational potential energy to find the speed of the rock as it left your hand:

mgh = 1/2mv^2

Where m is the mass of the rock, g is the acceleration due to gravity, h is the height difference between your hand and the Frisbee, and v is the speed of the rock as it left your hand. Solving for v:

v = √(2gh)

Substituting the given values:

v = √(2 * 9.8 * 2.8) = 8.18 m/s

Therefore, the speed of the rock as it left your hand was 8.18 m/s.

Answer 2

Answer: The rock's speed as it left your hand was 8.8 m/s.

Step-by-step explanation: The system is the rock and the Earth. The initial state is the rock at rest in your hand 2.8 m below the Frisbee. The final state is the rock hitting the Frisbee at a speed of 4.0 m/s.

Using conservation of energy, we know that the initial potential energy of the rock-Earth system is transformed into both kinetic energy and potential energy at its maximum height. Therefore, we can use the conservation of energy equation:

potential energy (initial) = kinetic energy (final) + potential energy (final)

mgh = 1/2mv^2 + mgh

where m is the mass of the rock, g is the acceleration due to gravity, h is the height that the rock has been raised, and v is the velocity of the rock.

We can solve for the initial velocity by rearranging the equation:

v = sqrt(2gh + v^2)

Plugging in the values, we get:

v = sqrt(2 * 9.81 * 2.8 + 4^2)

v ≈ 8.8 m/s

Therefore, the rock's speed as it left your hand was 8.8 m/s.