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.

Short answer

Question: Determine the coordinates where the emergency packet touches the ground and graph the path of the packet based on the given parametric equations x(t) = 80t and y(t) = -4.9t^2 + 3000. Answer: The emergency packet touches the ground at the coordinates (80 × √(3000/4.9), 0). To graph its path, plot the parametric equations x(t) = 80t and y(t) = -4.9t^2 + 3000 for t ≥ 0. The trajectory is a parabolic curve opening downwards with the starting point at (0, 3000) and touching the ground at the calculated landing point.

Step-by-step solution

Solve for time when packet touches the ground

To find when the packet touches the ground, set y(t) = 0 and solve for t: $$-4.9t^2 + 3000 = 0$$ First, isolate the t^2 term: $$t^2 = \frac{3000}{4.9}$$ Next, find the square root of both sides to solve for t: $$t = \sqrt{\frac{3000}{4.9}}$$

Calculate the x-coordinate of the landing point

Plug the value of t found in step 1 into the x(t) equation to find the x-coordinate: $$x = 80 \times \sqrt{\frac{3000}{4.9}}$$

Find the coordinates of the landing point

The coordinates of the landing point are given by (x, y) where x is calculated in step 2 and y = 0 (since the packet is on the ground): $$\text{Landing Point} = \left(80 \times \sqrt{\frac{3000}{4.9}}, 0 \right)$$

Graph the trajectory

To graph the trajectory of the packet, plot the parametric equations x(t) = 80t and y(t) = -4.9t^2 + 3000 for t ≥ 0. The trajectory will be a parabola opening downwards, starting from the point (0, 3000) and touching the ground at the landing point calculated in step 3.

Key concepts

Parametric Equations

In physics and mathematics, parametric equations are used to describe the trajectory or path of a moving object. In this exercise, the motion of the packet released from a plane is described using two separate equations: one for position in the horizontal direction ( x(t) = 80t d) and another for position in the vertical direction ( y(t) = -4.9t^2 + 3000 d). These equations are known as parametric equations because they use a parameter, in this case, time ( t d), to express the coordinates of the trajectory.

The horizontal equation, x = 80t d, tells us that the packet continues to move horizontally at a constant speed of 80 m/s, since there are no forces acting in that direction. The vertical equation, y = -4.9t^2 + 3000 d, describes how the packet falls due to gravity, with -4.9t^2 d representing the influence of gravity pulling the packet downwards, and 3000 d indicating the initial height from which it was dropped.

By using these two equations together, you can find the position of the packet at any time after it is released.

Graphing Trajectories

Graphing a trajectory involves plotting the path of a moving object to visualize its motion over time. The trajectory given in the exercise is plotted using the two parametric equations. The x-axis represents horizontal distance, while the y-axis represents height or vertical distance.

To graph the trajectory, you start at the initial point (0, 3000), which represents the position where the object begins its descent.

• The equation x = 80t d provides the horizontal position over time, starting at zero and continuing to increase as time progresses.
• The equation y = -4.9t^2 + 3000 d provides the vertical position, starting from 3000 meters and decreasing as the packet falls due to gravity.

The result is a downward-opening parabola, reflecting how the object ascends slightly due to initial motion, peaks, and then descends until it hits the ground.

Following the packet’s path visually helps to understand each phase of its movement, from release to landing.

Quadratic Equations

Quadratic equations are mathematical expressions of the form y = ax^2 + bx + c d. They appear frequently when describing parabolic motion, like the trajectory of the falling packet.

The vertical component of the packet's motion, y = -4.9t^2 + 3000 d, is a quadratic equation where a = -4.9 d and c = 3000 d (with b = 0 d because there is no linear t term). This equation is what gives the trajectory its characteristic parabolic shape.

Quadratic equations like this one can be solved for their roots to find specific events, such as the time when the packet hits the ground. By setting y d to zero, you can determine the time ( t d), which then becomes essential for calculating where exactly the packet lands in terms of horizontal distance. The solution involves isolating t^2 d and then calculating the square root, leading us to find the time it takes to reach the ground and use it in further calculations.