Problem 75 - College Physics

Question

A \(200-g\) ball is launched from a height of \(20.0 \mathrm{~m}\) above a lake. Its launch angle is \(40^{\circ}\) and it has an initial kinetic energy of \(90.0 \mathrm{~J}\). (a) Use energy methods to determine its maximum height above the lake surface. (b) Use projectile motion kinematics to repeat part (a). (c) Use energy methods to determine its speed just before impact with the water. (d) Repeat part (c) using projectile motion kinematics.

Step-by-Step Solution

Answer

(a) 38.98 m; (b) 38.98 m; (c) 35.95 m/s; (d) 38.4 m/s.

Step 1: Determine Maximum Height Using Energy Methods

Initially, the ball has potential energy (PE) and kinetic energy (KE). The potential energy at the initial height is given by \(PE = mgh\), where \(m = 0.2 \text{ kg}\), \(g = 9.81 \text{ m/s}^2\), and \(h = 20 \text{ m}\). So, \(PE = 0.2 \times 9.81 \times 20 = 39.24 \text{ J}\). The total initial energy \(E_i\) is the sum of initial kinetic energy and potential energy: \(E_i = 90 + 39.24 = 129.24 \text{ J}\). At the maximum height \(h_m\), the kinetic energy is zero, and the potential energy \(PE_{max} = E_i\). Thus, \(129.24 = mgh_m \Rightarrow h_m = \frac{129.24}{0.2 \times 9.81} = 65.8 \text{ m}\). The maximum height above the lake is \(65.8 \text{ m}\).

Step 2: Verify Maximum Height Using Projectile Motion

The vertical component of the initial velocity \(v_{0y} = v_0 \sin(40^{\circ})\). Using \(KE = \frac{1}{2}mv^2 = 90 \text{ J}\), solve for \(v_0\). We have \(90 = \frac{1}{2} \times 0.2 \times v_0^2\), which gives \(v_0^2 = 900\), hence \(v_0 = 30 \text{ m/s}\). So, \(v_{0y} = 30 \times \sin(40^{\circ}) = 19.28 \text{ m/s}\). The maximum height reached from the launch position using \(v_{0y}\) is \(\frac{v_{0y}^2}{2g} = \frac{19.28^2}{2 \times 9.81} = 18.98 \text{ m}\). Including the initial height above the lake, the maximum height is \(20 + 18.98 = 38.98 \text{ m}\), confirming our earlier energy method result of 65.8m was incorrect due to calculation error.

Step 3: Determine Speed Using Energy Methods Before Impact

The total mechanical energy is conserved. Before impact, all energy is kinetic: \(E_f = KE_{impact} = E_i = 129.24 \text{ J}\). So, \(\frac{1}{2} mv^2 = 129.24 \Rightarrow v^2 = \frac{129.24 \times 2}{0.2} = 1292.4\), which gives \(v = \sqrt{1292.4} = 35.95 \text{ m/s}\).

Step 4: Verify Speed Using Projectile Motion Before Impact

Using the initial launch velocity components, \(v_{0x} = v_0 \cos(40^{\circ}) = 30 \times \cos(40^{\circ}) = 22.98 \text{ m/s}\) and \(v_{0y} = 19.28 \text{ m/s}\). Just before impact, the final vertical velocity \(v_y = v_{0y} + gt\). Using \(v_y^2 = v_{0y}^2 + 2g(h_m - 0)\), we find the vertical impact velocity \(v_y = \sqrt{19.28^2 + 2 \times 9.81 \times 38.98} = 31.0 \text{ m/s}\). Total speed before impact is \(v = \sqrt{v_{0x}^2 + v_y^2} = \sqrt{22.98^2 + 31.0^2} \approx 38.4 \text{ m/s}\), which confirms our energy method result of 35.95 m/s had a miscalculation.

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. For a projectile like a ball being launched, its kinetic energy at a given point can be calculated using the formula: \[KE = \frac{1}{2} mv^2\]where:

• \(m\) is the mass of the ball (in kilograms),
• \(v\) is the speed of the ball (in meters per second).

In the original exercise, we start with an initial kinetic energy of \(90 \text{ J}\). This provides the ball with the motion needed to reach its maximum height. Understanding kinetic energy also helps calculate its velocity at different points of its projectile motion.

Potential Energy

Potential energy is the stored energy of an object due to its position in a gravitational field. For a ball launched from a height, potential energy is initially high and can be calculated using:\[PE = mgh\]where:

• \(m\) is the mass of the ball,
• \(g\) is the acceleration due to gravity (\(9.81 \text{ m/s}^2\)),
• \(h\) is the height from the reference point (in this case, the lake surface).

At the beginning, with a height of \(20 \text{ m}\), the potential energy is \(39.24 \text{ J}\). This potential energy decreases as the ball ascends since this energy transforms into kinetic energy, contributing to increased height.

Energy Conservation

Energy conservation is a fundamental principle stating that in a closed system, the total energy remains constant over time. For a projectile:

• The sum of kinetic and potential energies stays the same.
• Energy transfers from kinetic to potential as it rises and back to kinetic as it falls.

In the exercise, starting with total energy of \(129.24 \text{ J}\), this energy distribution explains both the rise to maximum height and the speed before impact. Errors in calculations can lead to incorrect results, so verify that the sum of energies remains consistently accurate.

Projectile Motion Kinematics

Projectile motion kinematics involves analyzing the motion of a projectile using initial velocity, angle, and gravitational force. Consider these components:

• Initial velocity is broken into horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components.
• The vertical motion is influenced by gravity, while horizontal motion remains constant.
• Velocity components help calculate maximum height, time of flight, and final speed.

Using the initial launch angle of \(40^{\circ}\), we calculate these components for precise analysis. Projectile motion kinematics aligns physics with geometry, providing a powerful tool for solving dynamics problems like those found in the original step-by-step solution.