Problem 48 - University Physics with Modern Physics

Question

A movie stuntwoman drops from a helicopter that is 30.0 m above the ground and moving with a constant velocity whose components are 10.0 m/s upward and 15.0 m/s horizontal and toward the south. Ignore air resistance. (a) Where on the ground (relative to the position of the helicopter when she drops) should the stuntwoman have placed foam mats to break her fall? (b) Draw \(x\)-\(t\), \(y\)-\(t\), \(v$$_x\)-\(t\), and \(v$$_y\)-\(t\) graphs of her motion.

Step-by-Step Solution

Answer

Place the mats 45.45 m south of the drop point. Graphs show constant horizontal motion and accelerated vertical motion.

Step 1: Determine time of fall

The initial vertical velocity component is given as \( v_{y0} = 10.0 \, \text{m/s} \) (upward) and the initial height is \( y_0 = 30.0 \, \text{m} \). Use the equation of motion for vertical displacement: \[ y = v_{y0}t + \frac{1}{2}gt^2 \]where \( g = -9.8 \, \text{m/s}^2 \) is the acceleration due to gravity (negative because it acts downward). We need to find \( t \) when \( y = 0 \), solving:\[ 0 = 30.0 + 10.0t - 4.9t^2 \]Using the quadratic formula, solve for \( t \). The roots are: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = -4.9 \), \( b = 10.0 \), \( c = 30.0 \). This gives us \( t \approx 3.03 \, \text{s} \).

Step 2: Horizontal displacement calculation

The horizontal velocity component \( v_{x0} \) is constant at \( 15.0 \, \text{m/s} \). The horizontal displacement \( x \) can be calculated using:\[ x = v_{x0}t \]Substitute \( t = 3.03 \, \text{s} \) from Step 1 into the equation:\[ x = 15.0 \times 3.03 = 45.45 \, \text{m} \].The foam mats should be placed \( 45.45 \, \text{m} \) south of the dropping point.

Step 3: Velocity and position graph description

1. **\( x-t \) graph**: A straight line with a slope of \( 15.0 \, \text{m/s} \), showing constant velocity at \( 15.0 \, \text{m/s} \). 2. **\( y-t \) graph**: A parabolic downward curve, starting at \( 30.0 \, \text{m} \), crossing \( y = 0 \) at \( t = 3.03 \, \text{s} \).3. **\( v_x-t \) graph**: A horizontal line indicating constant velocity at \( 15.0 \, \text{m/s} \).4. **\( v_y-t \) graph**: A straight line starting at \( 10.0 \, \text{m/s} \) and decreasing linearly, intersecting the \( x \)-axis at \( t \) when \( y eq 0 \), until \( -19.8 \, \text{m/s} \) at the time she hits the ground.

Key Concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause these motions. It is essential in understanding projectile motion, such as the scenario involving the stuntwoman leaping from a helicopter.
To predict where she lands, we need to consider her movements in both vertical and horizontal directions separately, which is a key feature of kinematics.

**Vertical Motion:**
The initial vertical velocity is given by the helicopter's upward speed, which is 10.0 m/s. Gravity will act against this motion, decelerating her ascent and then accelerating her descent. The kinematic equation \( y = v_{y0}t + \frac{1}{2}gt^2 \) helps calculate her vertical path. Here, the acceleration due to gravity \( g \) is -9.8 m/s². Setting the vertical displacement \( y \) to zero allows us to solve for time using the quadratic formula.

**Horizontal Motion:**
While gravity affects vertical motion, it doesn't influence the horizontal component. Thus, she travels horizontally at a steady 15.0 m/s. Calculating the distance she covers horizontally only involves the equation \( x = v_{x0}t \), highlighting the simplicity in horizontal kinematics.

Understanding these separate movements holds the key to predicting complex trajectories, which is why kinematics is a foundational concept in physics.

Quadratic Equations in Physics

In kinematics, quadratic equations often arise when dealing with objects moving under the influence of gravity. The classic projectile motion problem, like the stuntwoman's fall, uses a quadratic equation to determine the time of impact.

The equation:
\[ 0 = 30.0 + 10.0t - 4.9t^2 \]
comes from combining the formula for position \( y = v_{y0}t + \frac{1}{2}gt^2 \) and setting her altitude to zero when she hits the ground.

**Using the Quadratic Formula**

• To solve for \( t \), rearrange the equation into standard form \( ax^2 + bx + c = 0 \).
• Here, \( a = -4.9 \), \( b = 10.0 \), and \( c = 30.0 \).
• Use the quadratic formula:
  \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  to find the time when she reaches the ground.

Quadratic equations are not just limited to kinematics but extend to many physics problems. They help understand how time and position relate when objects undergo acceleration, particularly due to gravity.

Velocity-Time Graphs Analysis

Velocity-time graphs offer insightful visual representations of an object's velocity changes over time. They are particularly helpful when analyzing projectile motion.

**Understanding the Graphs**

• **Horizontal Velocity \(v_x - t \) Graph**:
  Given constant horizontal velocity of 15.0 m/s, the graph is a straight, horizontal line, indicating no change in speed horizontally over time.
• **Vertical Velocity \(v_y - t \) Graph**:
  This graph starts at 10.0 m/s and linearly decreases due to gravity's influence. When the stuntwoman hits the ground, \( v_y \) gets more negative, reaching -19.8 m/s, showing how her velocity changes from initially moving upward to accelerating downward.

Velocity-time graphs not only provide a snapshot of velocity changes but can also help calculate acceleration and total displacement. For instance, the area under these graphs can give the displacement over the time interval.

By studying these graphs, students can gain a deeper understanding of motion dynamics.