Solving Projectile Motion Using Energy: Videos & Practice Problems

Energy conservation is a powerful tool for solving projectile motion problems, allowing for quicker calculations of speeds and heights. For example, when a ball is thrown from a 30-meter building with an initial speed of 28 m/s, energy conservation can be used to find its speed just before hitting the ground. The equation combines kinetic energy and gravitational potential energy, leading to the result that the speed before impact is approximately 31.4 m/s. This method simplifies complex problems by eliminating the need for unknown launch angles.

Projectile Motion with Energy

Energy conservation is a powerful tool in solving problems related to projectile motion, allowing for quicker solutions to familiar scenarios. In this context, consider a ball thrown from the top of a 30-meter building with an initial speed of 28 meters per second. The goal is to determine the speed of the ball just before it hits the ground, using the principle of energy conservation rather than traditional projectile motion equations.

To analyze the situation, we define key points in the ball's trajectory: point A (launch), point B (maximum height), point C (initial height during descent), and point D (just before impact). The initial speed at point A is denoted as \( v_A = 28 \, \text{m/s} \), and we aim to find the speed at point D, \( v_D \).

Applying the energy conservation equation, we express the total mechanical energy at points A and D. The equation can be written as:

\[ K_A + U_A + W_{\text{non-conservative}} = K_D + U_D \]

Where:

• \( K \) represents kinetic energy, calculated as \( K = \frac{1}{2} mv^2 \)
• \( U \) represents gravitational potential energy, calculated as \( U = mgh \)
• \( W_{\text{non-conservative}} \) accounts for work done by non-conservative forces, which we will ignore in this case.

At point A, the kinetic energy is given by:

\[ K_A = \frac{1}{2} m v_A^2 = \frac{1}{2} m (28)^2 \]

The potential energy at point A, where the height \( h_A = 30 \, \text{m} \), is:

\[ U_A = mgh_A = mg(30) \]

At point D, just before hitting the ground, the kinetic energy is:

\[ K_D = \frac{1}{2} m v_D^2 \]

And the potential energy is zero since the height \( h_D = 0 \). Thus, the energy conservation equation simplifies to:

\[ \frac{1}{2} m (28)^2 + mg(30) = \frac{1}{2} m v_D^2 \]

We can cancel the mass \( m \) from all terms, leading to:

\[ \frac{1}{2} (28)^2 + g(30) = \frac{1}{2} v_D^2 \]

Substituting \( g = 9.8 \, \text{m/s}^2 \), we calculate:

\[ \frac{1}{2} (28)^2 + 9.8(30) = \frac{1}{2} v_D^2 \]

Calculating the left side:

\[ \frac{1}{2} (784) + 294 = \frac{1}{2} v_D^2 \]

This results in:

\[ 392 + 294 = \frac{1}{2} v_D^2 \]

Combining these gives:

\[ 686 = \frac{1}{2} v_D^2 \]

Multiplying both sides by 2 yields:

\[ 1372 = v_D^2 \]

Taking the square root provides the final speed:

\[ v_D = \sqrt{1372} \approx 37.0 \, \text{m/s} \]

This example illustrates how energy conservation simplifies the analysis of projectile motion, avoiding the complexities of unknown angles and allowing for a more straightforward calculation of the final speed just before impact.

Calculating Max Height with Energy Conservation

In this problem, we analyze the motion of a ball launched at a speed of 20 m/s and an angle of 37 degrees, focusing on calculating the maximum height using the principle of energy conservation rather than traditional projectile motion equations. The ball follows a parabolic trajectory, and we identify three key points: point A (the launch point), point B (the maximum height), and point C (the return to the initial height).

To find the maximum height at point B, we apply the energy conservation equation, which states that the total mechanical energy at point A equals the total mechanical energy at point B. The equation can be expressed as:

\[ K_A + U_A + W_{nc} = K_B + U_B \]

Here, \(K\) represents kinetic energy, \(U\) represents potential energy, and \(W_{nc}\) denotes work done by non-conservative forces. At point A, the ball has kinetic energy due to its initial velocity, while the potential energy is zero since we set the ground level as our reference point. There are no non-conservative forces acting on the ball, as we only consider gravitational forces.

At point B, the ball has kinetic energy due to its horizontal velocity and potential energy due to its height above the ground. The kinetic energy at point A can be calculated as:

\[ K_A = \frac{1}{2} m v_A^2 \]

At point B, the kinetic energy is given by:

\[ K_B = \frac{1}{2} m v_B^2 \]

Since the vertical component of the velocity at the peak (point B) is zero, the horizontal component remains constant throughout the motion. The horizontal velocity \(v_B\) can be determined using the cosine of the launch angle:

\[ v_B = v_A \cdot \cos(\theta) = 20 \cdot \cos(37^\circ) \approx 16 \text{ m/s} \]

Substituting these values into the energy conservation equation, we have:

\[ \frac{1}{2} m (20^2) = \frac{1}{2} m (16^2) + m g h_B \]

By canceling the mass \(m\) from all terms and multiplying the entire equation by 2 to eliminate the fractions, we simplify it to:

\[ 20^2 = 16^2 + 2gh_B \]

Rearranging this equation allows us to solve for the maximum height \(h_B\):

\[ h_B = \frac{20^2 - 16^2}{2g} \]

Substituting \(g \approx 9.8 \text{ m/s}^2\) into the equation, we find:

\[ h_B = \frac{400 - 256}{2 \cdot 9.8} \approx 7.35 \text{ meters} \]

This calculation reveals that the maximum height reached by the ball is approximately 7.35 meters, demonstrating the application of energy conservation principles in projectile motion analysis.

You are practicing jumping as far as you can. In one attempt, you run and leave the floor with 7 m/s directed at an unknown angle. What maximum height do you reach if your speed at that point is 5 m/s? Ignore air resistance.

• A. 1.22 m
• B. 1.73 m
• C. 2.44 m
• D. 0.67 m

Answer: (A)

Here’s what students ask on this topic:

How can energy conservation be used to solve projectile motion problems?

Energy conservation simplifies projectile motion problems by using the principle that the total mechanical energy (kinetic + potential) remains constant if only conservative forces (like gravity) are acting. For example, if a ball is thrown from a height, you can use the initial kinetic energy and gravitational potential energy to find the speed just before it hits the ground. The equation is:

$K_a+U_a=K_d+U_d$

where $K_a$ and $U_a$ are the initial kinetic and potential energies, and $K_d$ and $U_d$ are the final kinetic and potential energies. This method avoids complex calculations involving unknown angles.

What is the advantage of using energy conservation over motion equations in projectile motion?

Using energy conservation in projectile motion problems offers a significant advantage by simplifying the calculations. It eliminates the need to deal with unknown variables such as launch angles and time of flight. Instead, it focuses on the total mechanical energy, which remains constant. For instance, to find the speed of a projectile just before it hits the ground, you only need to consider the initial kinetic and potential energies and set them equal to the final kinetic energy. This approach is often quicker and more straightforward than using kinematic equations.

How do you calculate the speed of a projectile before it hits the ground using energy conservation?

To calculate the speed of a projectile before it hits the ground using energy conservation, follow these steps:

1. Identify the initial kinetic energy ($\frac{1}{2}mv$_a²) and gravitational potential energy (mgh_a).

2. Set up the energy conservation equation: $\frac{1}{2}mv$_a² + mgh_a = 12mv_d².

3. Cancel out the mass (m) from both sides.

4. Solve for the final speed (v_d): $v$_d = √(v_a² + 2gh_a).

For example, if a ball is thrown with an initial speed of 20 m/s from a height of 30 meters, the speed before it hits the ground is approximately 31.4 m/s.

What are the key points to consider when using energy conservation for projectile motion problems?

When using energy conservation for projectile motion problems, consider the following key points:

1. Identify the initial and final points of the motion.

2. Calculate the initial kinetic energy ($\frac{1}{2}mv$_a²) and gravitational potential energy (mgh_a).

3. Set up the energy conservation equation: $\frac{1}{2}mv$_a² + mgh_a = 12mv_d².

4. Cancel out the mass (m) if it appears in all terms.

5. Solve for the unknown variable, such as the final speed (v_d).

6. Ignore non-conservative forces like air resistance unless specified.

These steps help simplify the problem and avoid complex calculations involving unknown angles or time.

Can energy conservation be used to find the maximum height of a projectile?

Yes, energy conservation can be used to find the maximum height of a projectile. At the maximum height, the vertical component of the velocity is zero, so the kinetic energy is only due to the horizontal component. The energy conservation equation is:

$\frac{1}{2}mv$_a² + mgh_a = 12mv_h² + mgh_max

where v_a is the initial speed, h_a is the initial height, v_h is the horizontal speed at maximum height, and h_max is the maximum height. By solving for h_max, you can determine the maximum height reached by the projectile.