Problem 80 - Physics for Scientists and Engineers with Modern Physics

Question

A projectile is fired at an upward angle of \(48.0^{\circ}\) from the top of a 135 -m-high cliff with a speed of \(165 \mathrm{~m} / \mathrm{s}\). What will be its speed when it strikes the ground below? (Use conservation of energy.)

Step-by-Step Solution

Answer

The speed of the projectile when it hits the ground is approximately 172.9 m/s.

Step 1: Identify known values

We start with the known values from the problem: The initial velocity of the projectile, \(v_0 = 165\, \mathrm{m/s}\), the angle of launch, \(\theta = 48.0^\circ\), and the height of the cliff, \(h = 135\, \mathrm{m}\).

Step 2: Break down the initial velocity into components

The initial velocity has both horizontal \((v_{0x})\) and vertical \((v_{0y})\) components. We use trigonometry to find these components:\[ v_{0x} = v_0 \cdot \cos{\theta} = 165 \cdot \cos{48.0^\circ} \] \[ v_{0y} = v_0 \cdot \sin{\theta} = 165 \cdot \sin{48.0^\circ} \]

Step 3: Apply conservation of energy

According to the conservation of energy, the initial mechanical energy (sum of kinetic and potential energy at the top) equals the final mechanical energy (kinetic energy at the ground level since height is zero at the ground). The initial energy is:\[ E_i = \frac{1}{2}mv_0^2 + mgh \]The final energy is:\[ E_f = \frac{1}{2}mv_f^2 \]Setting \(E_i = E_f\), we have:\[ \frac{1}{2}mv_0^2 + mgh = \frac{1}{2}mv_f^2 \]

Step 4: Simplify the equation

Notice that the mass \(m\) cancels out from both sides of the equation. Therefore, \[ \frac{1}{2}v_0^2 + gh = \frac{1}{2}v_f^2 \]Re-arranging for \(v_f\), we get:\[ v_f = \sqrt{v_0^2 + 2gh} \]

Step 5: Calculate final speed

Substitute the known values into the equation from the previous step:\[ v_f = \sqrt{165^2 + 2 \times 9.8 \times 135} \]Calculating this results in:\[ v_f = \sqrt{27225 + 2646} = \sqrt{29871} \approx 172.9\, \mathrm{m/s} \]

Step 6: Conclusion

The speed of the projectile when it hits the ground is approximately \(172.9\, \mathrm{m/s}\).

Key Concepts

Conservation of Energy

The conservation of energy is a pivotal principle in physics, particularly in solving projectile problems. It asserts that energy cannot be created or destroyed, only transformed from one form to another. In the context of projectile motion, mechanical energy, which is the combination of potential energy (due to height) and kinetic energy (due to motion), remains constant if air resistance is ignored.

In this exercise, we start with a projectile launched from a high cliff. Initially, it possesses kinetic energy because of its speed and potential energy because of its height. As the projectile descends and strikes the ground, its height becomes zero, converting all its initial potential energy into kinetic energy. By equating the initial total mechanical energy to the final kinetic energy, we can calculate the projectile’s speed at impact. This approach wonderfully demonstrates how energy transformation can simplify complex motion problems.

Kinematics

Kinematics involves the study of motion without considering its causes. It's essential for analyzing projectile motion, as we break down the trajectory into horizontal and vertical components.

Initially, kinematics helps us resolve the projectile's speed into two components: horizontal (vx) and vertical (vy). These depend on the launch angle and initial speed, defined by trigonometric functions. Horizontal motion happens without acceleration (ignoring air resistance), while vertical motion is influenced by gravity. Understanding these separate components simplifies our approach by allowing us to tackle each direction independently before piecing together the overall motion description.

Physics Problem Solving

Successfully tackling physics problems involves a systematic approach to find the solution efficiently. First, it's critical to identify known values, such as initial velocity, launch angle, and height in projectile problems like this one.

Next, relying on the fundamental principles like conservation of energy and kinematics, we break down complex parts into simpler calculations – like resolving velocities into their components. Be mindful of assumptions in each model, such as ignoring air resistance or treating gravitational acceleration as constant. Lastly, reviewing the problem step-by-step to ensure consistency and correctness is vital to reaching an accurate solution.

Trigonometry in Physics

Trigonometry is indispensable in projectile calculations, particularly for determining velocity components in kinematics. Launch angles and direction manipulations heavily rely on trigonometric functions sine, cosine, and tangent.

In this exercise, we use trigonometry to find the horizontal and vertical velocity components from the initial speed. For a given angle \( \theta \), the horizontal component \( v_{0x} \) is \( v_0 \cdot \cos(\theta) \), and the vertical component \( v_{0y} \) is \( v_0 \cdot \sin(\theta) \). These calculations give us a detailed understanding of the initial trajectory's components, allowing us to apply further physics concepts in determining the projectile's complete motion and eventual impact speed.