Problem 29 - Physics for Scientists and Engineers with Modern Physics

Question

(1) A diver running 2.3 \(\mathrm{m} / \mathrm{s}\) dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. How high was the cliff and how far from its base did the diver hit the water?

Step-by-Step Solution

Answer

The cliff height is 44.1 meters, and the diver lands 6.9 meters from the cliff's base.

Step 1: Identify the Given Variables

We know that the horizontal speed of the diver is 2.3 m/s, and the time taken to reach the water is 3.0 seconds. Gravity's acceleration is 9.8 m/s².

Step 2: Determine Vertical Distance (Height)

To find the height of the cliff, use the formula for vertical motion: \[ h = \frac{1}{2}gt^2 \]Substitute the known values:\[ h = \frac{1}{2}(9.8 \text{ m/s}^2)(3.0 \text{ s})^2 \]\[ h = \frac{1}{2}(9.8)(9) \]\[ h = \frac{1}{2}(88.2) = 44.1 \text{ meters} \]So the height of the cliff is 44.1 meters.

Step 3: Determine Horizontal Distance

Use the formula for horizontal distance when the speed is constant:\[ d = vt \]Substitute the known values:\[ d = (2.3 \text{ m/s})(3.0 \text{ s}) \]\[ d = 6.9 \text{ meters} \]Thus, the diver hits the water 6.9 meters from the base of the cliff.

Key Concepts

Horizontal Motion in Projectile Motion

Projectile motion can be split into two distinct components: horizontal and vertical motion. Horizontal motion, in this context, refers to the movement of an object along a straight trajectory parallel to the ground.
When dealing with horizontal motion, a crucial assumption often holds true: the speed remains constant since there's usually no acceleration along the horizontal axis. In simpler terms, horizontal velocity stays the same as there are no opposing forces acting on it (assuming air resistance is negligible).
To solve problems regarding horizontal motion, the basic formula used is

• distance = speed × time, or
• \[d = vt\].

In the exercise with the diver, the horizontal speed was given as 2.3 m/s, and the duration was 3.0 seconds. By multiplying these values, the horizontal distance the diver traveled was calculated, showing the simplicity of calculating horizontal motion when variables are known.
This straightforward relationship is crucial for solving not just projectile problems but any situation where constant horizontal speed is a factor.

Vertical Motion in Projectile Motion

Vertical motion is distinct from horizontal motion because it involves acceleration due to gravity. In free fall scenarios like our diver example, gravity is the only force acting on the object vertically.
Gravity's acceleration is approximately 9.8 m/s² on Earth. This uniform acceleration enables us to use a specific formula to calculate the vertical distance, such as the height of the cliff in our example. The formula used is

• height = \(\frac{1}{2}gt^2\), where \(t\) is the time duration.

By substituting known values, we calculated the height of the cliff, showing how gravitational force consistently increases an object's velocity as time progresses during free fall. This aspect of physics clearly illustrates the effect of natural forces on an object's vertical motion and highlights the interconnected nature of physics concepts.

Physics Problem Solving Techniques

Solving physics problems such as projectile motion often involves a blend of breaking down the problem into components and using established formulas. Here are some steps that might help tackle such problems:

• Understand the Problem: Identify what is being asked. Break down the problem into horizontal and vertical components.
• Identify Known Variables: Gather all given data like speed, time, and acceleration.
• Choose the Right Formula: Select the correct formula for each component. Use \(d = vt\) for horizontal and \(h = \frac{1}{2}gt^2\) for vertical.
• Substitute the Values and Solve: Plug in the known values into the equations and solve for the unknowns.
• Check Your Work: Always verify your final answer to ensure the solution makes sense both mathematically and in a real-world context.

This systematic approach ensures accuracy and helps students fully grasp the complex interplay of physics in everyday scenarios. Understanding each part and applying the correct formulas can greatly enhance problem-solving skills in physics.