Problem 104 - Calculus Early Transcendentals

Question

A plane traveling horizontally at \(80 \mathrm{m} / \mathrm{s}\) over flat ground at an elevation of 3000 m releases an emergency packet. The trajectory of the packet is given by $$x=80 t, \quad y=-4.9 t^{2}+3000, \quad \text { for } t \geq 0$$ where the origin is the point on the ground directly beneath the plane at the moment of the release. Graph the trajectory of the packet and find the coordinates of the point where the packet lands.

Step-by-Step Solution

Answer

Answer: The coordinates of the landing point are (1960, 0).

Step 1: Find the time when the packet lands

We know that the packet lands when its vertical position \(y\) is zero. Thus, we need to find \(t\) that satisfies the equation: $$ -4.9t^2 + 3000 = 0 $$ To solve this equation, divide both sides by -4.9: $$ t^2 = \frac{3000}{4.9} $$ Now, take the square root of both sides: $$ t = \sqrt{\frac{3000}{4.9}} $$ Compute the value of \(t\): $$ t \approx 24.5 \text{ seconds} $$

Step 2: Determine the packet's horizontal position when it lands

Now that we know the time that the packet lands, we can compute its horizontal position using the equation \(x=80t\). Substitute the value of \(t\) found in Step 1: $$ x = 80 (24.5) $$ Compute the value of \(x\): $$ x \approx 1960 \text{ meters} $$

Step 3: Find the coordinates of the point where the packet lands

Based on the results from Steps 1 and 2, the packet lands at time \(t \approx 24.5\) seconds with horizontal position \(x\approx 1960\) meters. Since we determined that the packet landed when \(y=0\), the coordinates of the landing point are: $$ (x, y) = (1960, 0) $$ So, the packet lands at the point \((1960, 0)\).

Key Concepts

Quadratic Equations

Understanding quadratic equations is crucial when analyzing motions like that of a projectile released from a plane. A quadratic equation can be identified by its standard form \( ax^2 + bx + c = 0 \), where \( a \) is not zero, and \( x \) represents an unknown variable.

In our exercise, the vertical motion of the emergency packet is modeled by the equation \( y=-4.9t^2+3000 \), which is a quadratic equation where \( y \) is the vertical height at any time \( t \). The fact that \( y \) becomes zero when the packet hits the ground provides the necessary condition to solve for time using the quadratic formula or by isolating \( t^2 \) and then taking the square root, as shown in the provided solution.

Key Features of Quadratic Functions

• They have a parabolic shape, opening upwards or downwards.
• The coefficient in front of \( t^2 \) determines the direction of the parabola.
• The maximum or minimum value (vertex) of the parabola represents the highest or lowest point in the physical trajectory of the projectile.
• The roots of the quadratic equation (when \( y=0 \)) give important points of the trajectory, such as the landing point.

Quadratic functions are a cornerstone in the interpretation of projectile motion and are widely used in various physics problems.

Graphing Motion

Graphing motion aids in visualizing the otherwise abstract equations and numbers. In our scenario, where a packet is dropped from a plane, a graph of its trajectory would show how the packet travels through the air over time.

Using the horizontal equation \( x=80t \) and the vertical equation \( y=-4.9t^2 +3000 \) allows us to create a two-dimensional graph that represents the packet’s path. The horizontal axis of this graph generally represents time (\( t \) in seconds), while the vertical axis represents height (\( y \) in meters).

How to Graph Projectile Motion

• Plot the horizontal and vertical positions as functions of time to show the path of the projectile.
• Draw the curve representing the quadratic function of vertical motion to see how the height changes over time.
• Use the horizontal motion equation to determine the relationship between distance traveled and time.
• The intersection with the horizontal axis on the motion graph signifies when the projectile lands.

Graphing both the vertical and horizontal components on the same set of axes can illustrate the full trajectory of the projectile, providing valuable insights regarding when and where it will land.

Kinematic Equations

Kinematic equations describe the motion of objects without considering the forces that cause the motion. These equations are essential for solving problems like the projectile exercise we are discussing.

Projectile motion is a prime example of kinematics in action. The projectile (in this case, the emergency packet) moves in two dimensions: horizontally with a constant velocity and vertically under the acceleration due to gravity. Here are the kinematic equations relevant to projectile motion:

• \( y = y_0 + v_{0y}t + \frac{1}{2}at^2 \) (Vertical motion)
• \( x = x_0 + v_{0x}t \) (Horizontal motion)

In the given exercise, \( -4.9t^2 + 3000 \) represents the vertical displacement \( y \) as a function of time \( t \) and the acceleration due to gravity (approximated as \( -9.8 \,\text{m/s}^2 \) divided by 2). No initial vertical velocity is given, suggesting the packet was released without being thrown upwards or downwards.

The horizontal motion \( x = 80t \) assumes a constant horizontal velocity along the flight path. There is no horizontal acceleration, which simplifies this part of the motion.

Kinematic equations let us predict the position and velocity of an object at any given time. For students studying physics, mastering these equations is invaluable for analyzing motion in a clear and methodical manner.