Question

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

Short answer

Release the package 2856 meters before the target.

Step-by-step solution

Understand the Equations

The trajectory of the package is given by the equations \( x = 100t \) and \( y = -4.9t^2 + 4000 \). Here, \( x \) gives the horizontal position of the package, and \( y \) gives the vertical position above the ground.

Determine When the Package Hits the Ground

The package hits the ground when \( y = 0 \). We set the vertical equation to zero: \(-4.9t^2 + 4000 = 0 \). Solve for \( t \) by rearranging the equation to \( 4.9t^2 = 4000 \).

Solve for Time of Impact

To find \( t \), divide both sides of the equation by 4.9: \( t^2 = \frac{4000}{4.9} \). Then take the square root: \( t = \sqrt{\frac{4000}{4.9}} \).

Calculate the Time

Calculate \( t = \sqrt{816.3265} \approx 28.56 \) seconds as the time when the package reaches the ground.

Calculate Horizontal Distance Before Release

For the horizontal distance, use \( x = 100t \). Substitute \( t = 28.56 \) to get \( x = 100 \times 28.56 = 2856 \) meters.

Determine Package Release Point

The package should be released 2856 meters before reaching the target to hit the target on the ground.

Key concepts

Understanding Horizontal Motion

In projectile motion, horizontal motion occurs without any acceleration if air resistance is negligible. For the emergency package dropped by the airplane, the horizontal position is given by the equation \( x = 100t \). This equation tells us that the package travels horizontally at a velocity of \( 100 \) meters per second.
This uniform motion means that for every second the package travel, it covers an additional 100 meters horizontally.

• The plane's speed defines how far the package goes along the ground during the time it falls.
• Horizontal distance traveled can be calculated readily with the time of flight.

Since the equation is a simple linear relation with time, by knowing the time it takes to hit the ground, we can compute exactly how far the package will move horizontally.

Vertical Motion Explained

Vertical motion of projectiles is quite different as it involves gravity. In this exercise, the vertical motion is defined by the equation \( y = -4.9t^2 + 4000 \). Here, \( y \) represents the height of the package above the ground.
At any time \( t \), we can determine the vertical position of the package using this equation. The term \(-4.9t^2\) comes from the gravitational acceleration. This is because in the context of projectile motion, the acceleration due to gravity is approximately \(9.8 \) m/s², but as we're considering the downward force, it's negative.

• The initial height is 4000 meters, the height of the plane.
• Gravity causes the vertical position to change as \(-4.9t^2\).

As time progresses, the package will drop towards the ground, accelerating downwards due to gravity.

Calculating Time of Flight

The time of flight refers to the time taken for the object to travel through the air before it hits the ground. For the package, it starts falling from the plane when \( t = 0 \) until it hits the ground when \( y = 0 \). To find when this happens, we solve the equation \(-4.9t^2 + 4000 = 0\).
Rearranging gives \(4.9t^2 = 4000\).
Solving for \(t\) involves division and then taking the square root:

• Divide both sides by 4.9, which results in \(t^2 = \frac{4000}{4.9}\).
• Taking the square root gives \(t ≈ 28.56\) seconds.

Thus, the package remains in the air for approximately 28.56 seconds before reaching the ground.

Impact Calculation and Targeting

The impact calculation is crucial for determining where the package will land. To ensure the package hits the intended target, we must calculate how far in advance it should be released using its horizontal motion. Given the horizontal motion equation \( x = 100t \), and knowing the time of flight is \( t = 28.56 \) seconds:

• Substitute \( t = 28.56 \) into the horizontal equation.
• This calculation gives \( x = 100 \times 28.56 = 2856 \) meters.

Hence, to ensure the package lands on target, it needs to be released 2856 meters before reaching the point directly over the target on the ground. This calculation provides exact release timing for precise delivery.