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Chapter 3: Problem 48

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.

Short Answer

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

Step by step solution

01

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 \( t \approx 3.03 \, \text{s} \).
02

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.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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/s虏. 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.

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Most popular questions from this chapter

A jet plane is flying at a constant altitude. At time \(t_1\) = 0, it has components of velocity \(v_x\) = 90 m/s, \(v_y\) = 110 m/s. At time \(t_2\) = 30.0 s, the components are \(v_x\) = -170 m/s, \(v_y\) = 40 m/s. (a) Sketch the velocity vectors at \(t_1\) and \(t_2\). How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

If \(\vec{r} = bt^2\hat{\imath} + ct^3\hat{\jmath}\), where \(b\) and \(c\) are positive constants, when does the velocity vector make an angle of 45.0\(^\circ\) with the \(x\)- and \(y\)-axes?

Two students are canoeing on a river. While heading upstream, they accidentally drop an empty bottle overboard. They then continue paddling for 60 minutes, reaching a point 2.0 km farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they turn around and head downstream. They catch up with and retrieve the bottle (which has been moving along with the current) 5.0 km downstream from the turnaround point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

Henrietta is jogging on the sidewalk at 3.05 m/s on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is 38.0 m above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally 9.00 s after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

CALC The position of a squirrel running in a park is given by \(\vec{r}= [(0.280 m/s)t + (0.0360 m/s^2)t^2] \hat{\imath}+(0.0190 m/s^3)t^3\hat{\jmath}\). (a) What are \(v_x(t)\) and \(v_y(t)\), the \(x\)- and \(y\)-components of the velocity of the squirrel, as functions of time? (b) At \(t\) = 5.00 s, how far is the squirrel from its initial position? (c) At \(t\) = 5.00 s, what are the magnitude and direction of the squirrel's velocity?
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