Problem 105 - Calculus Early Transcendentals

Question

A plane traveling horizontally at \(100 \mathrm{m} / \mathrm{s}\) over flat ground at an elevation of \(4000 \mathrm{m}\) must drop an emergency packet on a target on the ground. The trajectory of the packet is given by $$x=100 t, \quad y=-4.9 t^{2}+4000, \quad \text { for } t \geq 0,$$ where the origin is the point on the ground directly beneath the plane at the moment of the release. How many horizontal meters before the target should the packet be released in order to hit the target?

Step-by-Step Solution

Answer

Answer: The packet should be released 2860 horizontal meters before the target.

Step 1: Solve for the time when the packet reaches the ground.

To find the time when the packet reaches the ground, we need to set y = 0 and solve for t: $$0 = -4.9t^2 + 4000$$

Step 2: Solve the quadratic equation for t.

To find the time value, we need to solve the quadratic equation from Step 1: $$4.9t^2 = 4000$$ Divide both sides by 4.9: $$t^2 = \frac{4000}{4.9}$$ Now take the square root of both sides: $$t = \sqrt{\frac{4000}{4.9}}$$

Step 3: Calculate the time value.

Now, we can calculate the actual time value: $$t = \sqrt{\frac{4000}{4.9}} \approx 28.6\,\text{seconds}$$ The packet will take approximately 28.6 seconds to reach the ground.

Step 4: Calculate the horizontal distance before the target.

Now that we have the time value, we can use it to calculate the corresponding horizontal distance x. Remember that x = 100t. So, plug in the value of t we found: $$x = 100 \times 28.6$$

Step 5: Calculate the horizontal distance.

Finally, calculate the horizontal distance: $$x = 100 \times 28.6 = 2860\,\text{meters}$$ So, the packet should be released 2860 horizontal meters before the target to hit the target.

Key Concepts

Quadratic Equations in Projectile Motion

Quadratic equations are a key aspect when dealing with projectile motion problems. They appear frequently due to the relationship between time, velocity, and distance in both the horizontal and vertical directions.
In the exercise at hand, the vertical motion of the packet is modeled by the equation:

• \( y = -4.9t^2 + 4000 \)

This is a quadratic equation because it has the form \( y = ax^2 + bx + c \). In this case, it describes the height \( y \) of the packet as a function of time \( t \).

• The term \( -4.9t^2 \) represents the acceleration due to gravity affecting the packet as it falls, assuming the units are meters and seconds and gravity is \( 9.8\, \text{m/s}^2 \).
• The term \( 4000 \) is the initial height from which the packet is dropped.

To solve for the time \( t \) when the packet reaches the ground (\( y=0 \)), substitute \( y = 0 \) and solve the quadratic equation. This is crucial as it allows us to determine how long it takes for the packet to hit the ground, thereby enabling the calculation of horizontal distance.

Calculating Horizontal Distance

The horizontal distance covered in projectile motion is straightforward due to the constant horizontal speed. In our exercise, the horizontal distance \( x \) is given by:

• \( x = 100t \)

This formula comes from the fact that the horizontal velocity is \( 100 \, \text{m/s} \). Since there is no horizontal acceleration (assuming no air resistance), the distance traveled horizontally is simply the product of velocity and time.
To determine how many meters before the target the packet should be released, we first found \( t = 28.6 \, \text{seconds} \) from solving the quadratic for the vertical motion. Then plug this time into the equation for horizontal distance:

• \( x = 100 \times 28.6 = 2860 \, \text{meters} \)

Thus, the packet must be released 2860 meters away from the target to ensure it hits the correct spot.

Understanding Free Fall

Free fall describes the motion of an object under the influence of gravity alone, without any initial upward velocity. In projectiles, like in our problem, the free fall part pertains solely to the vertical motion. The projectile -- here, the packet -- is influenced by gravity, causing it to accelerate downward at \( 9.8 \, \text{m/s}^2 \).

• The equation \( y = -4.9t^2 + 4000 \) captures this free-fall behavior.
• The \( -4.9 \) coefficient is derived from half of the gravitational acceleration, \( -9.8/2 \), which is how standard physics equations represent the effect of gravity when squaring time \( t^2 \).

The packet lacks any propulsive force downward, making free fall the sole factor in its descent. Understanding this concept helps break down the problem into manageable parts: isolating what effects gravity has and knowing to solve for time and distance using those isolated portions. Free fall is crucial in predicting how long and how far the packet will travel vertically, thus influencing the horizontal distance calculation as well.