Airplane Suitcase
Projectile Motion

The Question

An airplane is flying with a velocity of $90.0 \text{ m/s}$ at an angle of $23.0^\circ$ above the horizontal. When the plane is $114 \text{ m}$ directly above a dog that is standing on level ground, a suitcase drops out of luggage compartment. How far from the dog will the suitcase land? You can ignore air resistance.

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Introduction

Projectile motion problems involving moving vehicles are classic physics staples. Whether it's a helicopters releasing relief supplies, or, in this case, an airplane dropping a suitcase, the underlying physics remains the same.

The key to solving this problem is understanding a fundamental principle of inertia: when an object is dropped from a moving vehicle, it initially shares the exact same velocity (speed and direction) as the vehicle.

In this guide, we will break down this 2D kinematics problem step-by-step, review the necessary formulas, and show you exactly how to find where the suitcase lands relative to the dog.

Concept Explanation: 2D Projectile Motion

To solve this, we must use the principles of two-dimensional kinematics. The most important rule to remember is that horizontal motion and vertical motion are completely independent of each other.

• Horizontal Motion ($x$-axis): Because we are ignoring air resistance, there are no forces acting horizontally on the suitcase once it drops. Therefore, horizontal velocity is constant ($a_x = 0$).
• Vertical Motion ($y$-axis): The suitcase is subject to the constant downward acceleration of gravity ($g \approx 9.8 \text{ m/s}^2$).

The Kinematic Equations

We will rely on the following standard kinematic equation for position:

$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$ $x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$

Since $a_x = 0$, the horizontal equation simplifies to: $x = x_0 + v_{0x}t$

Step-by-Step Solution

Step 1: Identify the Known Variables and Setup the Coordinate System

Let's place the origin $(0,0)$ at the location of the dog on the ground.

• Initial velocity of the plane (and suitcase), $v_0 = 90.0 \text{ m/s}$
• Launch angle, $\theta = 23.0^\circ$ (above horizontal)
• Initial vertical position, $y_0 = 114 \text{ m}$
• Final vertical position, $y = 0 \text{ m}$ (hitting the ground)
• Initial horizontal position, $x_0 = 0 \text{ m}$ (directly above the dog)
• Vertical acceleration, $a_y = -g = -9.8 \text{ m/s}^2$

Step 2: Resolve Initial Velocity into Components

The suitcase doesn't just fall straight down; it is launched upwards and forwards. We must find the $x$ and $y$ components of the initial velocity using trigonometry.

Horizontal initial velocity ($v_{0x}$): $v_{0x} = v_0 \cos(\theta)$ $v_{0x} = 90.0 \cos(23.0^\circ)$ $v_{0x} \approx 90.0 \times 0.9205 \approx 82.845 \text{ m/s}$

Vertical initial velocity ($v_{0y}$): $v_{0y} = v_0 \sin(\theta)$ $v_{0y} = 90.0 \sin(23.0^\circ)$ $v_{0y} \approx 90.0 \times 0.3907 \approx 35.166 \text{ m/s}$ (Note: This is positive because the plane is angled upwards).

Step 3: Find the Time of Flight ($t$)

We use the vertical motion equation to find how long the suitcase is in the air.

$y = y_0 + v_{0y}t - \frac{1}{2}gt^2$ $0 = 114 + 35.166t - \frac{1}{2}(9.8)t^2$ $0 = 114 + 35.166t - 4.9t^2$

To solve for $t$, we rearrange this into standard quadratic form ($at^2 + bt + c = 0$): $-4.9t^2 + 35.166t + 114 = 0$

Now, apply the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $t = \frac{-35.166 \pm \sqrt{(35.166)^2 - 4(-4.9)(114)}}{2(-4.9)}$ $t = \frac{-35.166 \pm \sqrt{1236.65 + 2234.4}}{-9.8}$ $t = \frac{-35.166 \pm \sqrt{3471.05}}{-9.8}$ $t = \frac{-35.166 \pm 58.916}{-9.8}$

This gives us two possible times:

1. $t = \frac{23.75}{-9.8} \approx -2.42 \text{ s}$ (We discard negative time in this context)
2. $t = \frac{-94.08}{-9.8} \approx 9.600 \text{ s}$

The suitcase is in the air for $9.60 \text{ s}$.

Step 4: Calculate the Horizontal Distance

Now that we have the time of flight, we can find out how far horizontally the suitcase travels from the drop point. Since the dog is directly below the drop point, this horizontal distance ($x$) is exactly how far away from the dog it will land.

$x = x_0 + v_{0x}t$ $x = 0 + (82.845 \text{ m/s})(9.600 \text{ s})$ $x \approx 795.31 \text{ m}$

Rounding to three significant figures (matching our initial given values of $90.0$, $23.0$, and $114$), we get $795 \text{ m}$.

Final Answer

The suitcase will land $795 \text{ m}$ away from the dog.

Common Mistakes

• Assuming initial vertical velocity is zero: A very common mistake is assuming $v_{0y} = 0$. Because the plane is flying at an upward angle of $23^\circ$, the suitcase also has an initial upward velocity. You must calculate the $v_{0y}$ component.
• Sign errors with gravity: If you set upwards as positive (which we did here by making $v_{0y}$ positive), gravity must be negative ($a_y = -9.8 \text{ m/s}^2$). Mixing up signs in the quadratic equation will lead to incorrect times.
• Rounding too early: Keep intermediate values (like $v_{0x}$ and $v_{0y}$) to several decimal places during the calculation. Only round to standard significant figures at the very last step.

Practice Questions

Want to test your skills? Try these variations:

1. Horizontal Drop: An airplane flying completely horizontally at $80 \text{ m/s}$ drops a package from a height of $500 \text{ m}$. How far horizontally does the package travel before hitting the ground?
2. Downward Angle: A helicopter is flying downward at an angle of $15^\circ$ below the horizontal with a speed of $45 \text{ m/s}$. If it drops a flare from $200 \text{ m}$ high, how long does it take to hit the ground?
3. Finding Velocity: Using the original suitcase problem above, what is the magnitude of the suitcase's velocity just before it hits the ground?

FAQs

Q: Why does the suitcase have an initial velocity if it was "dropped"? A: In physics, an object inside a moving vehicle shares the vehicle's velocity due to inertia. When released, it doesn't instantly stop; it continues forward (and in this case, upward) at the plane's exact velocity at the moment of release.

Q: Why do we ignore air resistance in these problems? A: Including air resistance (drag) makes the mathematics significantly more complex, requiring differential equations. In introductory physics, we ignore it to focus on understanding the fundamental principles of gravity and independent 2D motion.

Q: Does the mass of the suitcase affect how far it lands? A: No. In the absence of air resistance, all objects fall at the same rate regardless of their mass due to gravity. The mass variable cancels out in kinematic equations.

Q: What if the airplane was flying horizontally instead of at an angle? A: If the plane were flying horizontally, the initial vertical velocity ($v_{0y}$) would be exactly zero. The initial horizontal velocity ($v_{0x}$) would equal the full speed of the plane. The math becomes much easier because the $y$-equation simplifies to $y = y_0 - \frac{1}{2}gt^2$.

Conclusion

Solving projectile motion problems where the object is released from a moving vehicle requires careful attention to the initial conditions. By meticulously breaking the initial velocity into horizontal and vertical components, and remembering that horizontal and vertical motions are independent, you can confidently solve any variation of this classic physics problem. Keep practicing those kinematic equations!