Question

A \(0.26-\mathrm{kg}\) rock is thrown vertically upward from the top of a cliff that is \(32 \mathrm{~m}\) high. When it hits the ground at the base of the cliff, the rock has a speed of \(29 \mathrm{~m} / \mathrm{s}\). Assuming that air resistance can be ignored, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

Short answer

Initial speed is approximately 7.49 m/s and the greatest height from the base is 34.86 m.

Step-by-step solution

Understanding the Problem

We need to find the initial speed of a rock and the greatest height from the base of a cliff. The rock is thrown vertically upward, falls to the ground, covering a height difference of 32 m and reaching a final speed of 29 m/s.

Use Energy Conservation for Initial Speed

When the rock is thrown up, it has some initial kinetic energy and potential energy which, at the lowest point (ground level), get converted to kinetic energy. Using the conservation of mechanical energy: \[ \frac{1}{2} m v_i^2 + mgh = \frac{1}{2} m v_f^2 \] where \(m = 0.26 \text{ kg}\), \(g = 9.81 \text{ m/s}^2\), \(h = 32 \text{ m}\), and \(v_f = 29 \text{ m/s}\).

Solve for Initial Velocity \(v_i\)

Plug values into the energy conservation equation \[ \frac{1}{2} \times 0.26 \times v_i^2 + 0.26 \times 9.81 \times 32 = \frac{1}{2} \times 0.26 \times 29^2 \] Simplifying, we find \(v_i^2\): \[ \frac{1}{2} v_i^2 + 81.468 = 109.51 \] \[ \frac{1}{2} v_i^2 = 28.042 \] \[ v_i^2 = 56.084 \] \[ v_i = \sqrt{56.084} \approx 7.49 \text{ m/s} \]

Determine the Greatest Height

For the greatest height calculation, consider that at the peak the velocity is 0. Start with initial velocity and calculate the height gained from there. \[ v_i^2 = 2gh_{max} \] \[ (7.49)^2 = 2 \times 9.81 \times h_{max} \] \[ h_{max} = \frac{7.49^2}{2 \times 9.81} = 2.86 \text{ m} \]

Calculate Total Height from Base

Add the maximum height above the cliff to the height of the cliff: \(32 \text{ m} + 2.86 \text{ m} = 34.86 \text{ m}\). Thus, the greatest height the rock travels from the base is 34.86 m.

Key concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. The faster an object moves, the more kinetic energy it has. For an object with mass, such as our rock, kinetic energy is calculated using the formula: \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

When the rock is thrown upwards, it gains initial kinetic energy, which helps determine how high it will ascend. Because the rock is subject to gravity, this energy will subsequently convert into potential energy on its way up. As it rises, the rock's speed decreases, decreasing its kinetic energy, until at its peak, the velocity is zero and kinetic energy is minimal.

Potential Energy

Potential energy, specifically gravitational potential energy, is the energy stored by an object due to its position relative to the ground. For our rock, this energy is calculated using \( PE = mgh \), where \( m \) represents mass, \( g \) the gravitational acceleration (\( 9.81 \text{ m/s}^2 \)), and \( h \) the height.

In this scenario, the rock's potential energy is highest at the greatest height from the base of the cliff. Initially, as the rock leaves the top of the cliff, it already has a significant amount of potential energy due to its height above the ground. As the rock moves upward, the height relative to the ground increases even further, leading to increased potential energy.

Initial Velocity Calculation

The initial velocity of an object is the speed at which it begins its motion. Calculating this involves energy considerations, especially in a setting where mechanical energy conservation applies.

To find the initial velocity of the rock, we set up an equation based on the conservation of energy, which equates the initial total energy of the rock to its energy when it reaches the ground. This means using the energy conservation equation: \( \frac{1}{2} mv_i^2 + mgh = \frac{1}{2} mv_f^2 \).

By inserting the given values for mass, gravitational acceleration, height, and final velocity, we can solve for \( v_i \), which gives the initial velocity of approximately 7.49 m/s. This calculation involves assessing how energy transitions from potential to kinetic form through changes in height and speed.

Energy Conservation Equation

The energy conservation equation is a crucial tool in physics for solving problems involving motion and forces. It states that the total mechanical energy of a system remains constant if only conservative forces, like gravity, act upon it. Mathematically, this is expressed as: \( \text{Initial kinetic energy} + \text{Initial potential energy} = \text{Final kinetic energy} + \text{Final potential energy} \).

In this rock scenario, the energy conservation equation simplifies analysis by acknowledging that as the rock moves, no energy is lost to air resistance, and thus, the initial and final energies balance.

• The rock gains potential energy as it goes higher while kinetic energy decreases.
• Upon descent, potential energy converts back into kinetic energy, speeding up the rock.

This equation helps us determine features like the initial speed and greatest height reached, linking basic principles of energy conversion and mechanics.