Problem 54 - College Physics

Question

A good-guy stuntman is being chased by bad guys on a building's level roof. He comes to the edge and is to jump to the level roof of a lower building \(4.0 \mathrm{~m}\) below and \(5.0 \mathrm{~m}\) away. What is the minimum launch speed the stuntman needs to complete the jump? (Landing on the edge is assumed complete.)

Step-by-Step Solution

Answer

The minimum launch speed required is approximately 5.54 m/s.

Step 1: Understanding the Problem

The stuntman needs to jump from one building to another that is 4.0 meters lower and 5.0 meters away. We need to find the minimum horizontal velocity he needs to jump to safely land on the next building.

Step 2: Identify the Known Variables

The vertical distance (height) the stuntman will fall is 4.0 meters downward, and the horizontal distance (range) to be covered is 5.0 meters. The initial vertical velocity is 0 since he jumps horizontally.

Step 3: Calculate Time of Flight

First, we need to calculate the time it takes for the stuntman to fall 4.0 meters. We use the equation for vertical motion: \[ h = \frac{1}{2}gt^2 \]where \( h = 4.0 \text{ m} \) and \( g = 9.8 \text{ m/s}^2 \). Solving for \( t \):\[ 4.0 = \frac{1}{2}(9.8)t^2 \]\[ t^2 = \frac{4.0}{4.9} \approx 0.816 \]\[ t \approx \sqrt{0.816} \approx 0.903 \text{ seconds} \].

Step 4: Calculate Minimum Launch Speed

Now that we know the time of flight is approximately 0.903 seconds, we find the minimum launch speed using horizontal motion:\[ d = v_x t \]where \( d = 5.0 \text{ m} \) is the horizontal distance and \( t = 0.903 \text{ s} \). Solving for \( v_x \):\[ 5.0 = v_x (0.903) \]\[ v_x = \frac{5.0}{0.903} \approx 5.54 \text{ m/s} \].

Step 5: Summarize the Solution

The stuntman needs a minimum horizontal launch speed of approximately 5.54 meters per second to safely jump to the other building.

Key Concepts

Horizontal Launch Speed

When discussing projectile motion, horizontal launch speed is critical. When an object is launched horizontally, it starts with an initial speed in the horizontal direction. This speed doesn't change, as there are no forces acting horizontally to accelerate or decelerate the object during flight. To find the horizontal launch speed required for a stuntman to safely cross a gap, you consider the distance between the initial position and the landing target. Using the equation \( d = v_x \cdot t \), where \( d \) is the horizontal distance, \( v_x \) is the horizontal velocity, and \( t \) is time of flight, we calculate the speed needed. - In our stuntman's case, the gap measures 5.0 meters. - With the time of flight known, we divided the required distance by this time to find the necessary horizontal speed: \( v_x = \frac{5.0}{0.903} \approx 5.54 \text{ m/s} \). This calculation tells us the exact speed the stuntman needs to maintain horizontally to make a successful jump.

Time of Flight Calculation

In physics, calculating the time of flight for an object projected horizontally is essential. It determines how long the object will be in the air. This time is influenced only by vertical motion, not horizontal motion. The time of flight depends on how far down the object falls, due to gravity, which accelerates it downward. For our stuntman example, the equation \( h = \frac{1}{2}gt^2 \) helps calculate the time, \( t \), given \( h \), the vertical height, and \( g \), the acceleration due to gravity. - The vertical fall was 4.0 meters in this scenario. - Solving \( 4.0 = \frac{1}{2}(9.8)t^2 \), we find \( t \approx 0.903 \text{ seconds} \). Thus, the stuntman remains airborne for roughly 0.903 seconds as he makes his jump, a vital number for assessing necessary speeds and distances.

Vertical Motion Equation

The vertical motion of an object under gravity is described by specific equations that consider the initial position, velocity, and the constant acceleration due to gravity. In the context of horizontal projectile motion problems, the initial vertical velocity is often zero because the launch is purely horizontal. For our exercise, calculating vertical displacement uses the formula \( h = \frac{1}{2}gt^2 \). Here's what it incorporates: - \( h \) is the vertical distance fallen, which is 4.0 m in our case. - \( g \), the gravitational acceleration, is approximately 9.8 m/s². - \( t \) is the time taken to fall that distance. Knowing these values, we plug them into the equation to solve for \( t \). This is a foundational method in physics applications dealing with projects involving drops or falls over a known vertical distance. It highlights how gravity influences the vertical path and timing of a projectile.