Projectile Motion: Finding Horizontal Distance in Package Drop

Physics Motion 11th-12th Grade
PROBLEM
A rescue pilot wishes to drop a package of emergency supplies so that it lands as close as possible to a target. If the plane travels with a velocity of 81.00 m/s and is flying 125 m above the target, how far away (horizontally) from the target must the rescue pilot drop the package?

Visualizing the Problem

A rescue pilot wishes to drop a package of emergency supplies so that it lands as close as possible to a target. If...

What This Problem Teaches

  • Understanding projectile motion as independent horizontal and vertical components
  • Applying kinematic equations to find fall time from height
  • Recognizing that horizontal velocity remains constant during free fall
  • Connecting physics concepts to real-world rescue and delivery scenarios
  • Using the relationship between distance, velocity, and time in two dimensions

Solution: Method 1 — Independent Motion Analysis

The key insight is that horizontal and vertical motions are completely independent in projectile motion. We'll find the fall time first, then calculate how far the package travels horizontally during that time.

Step 1 — Identify the initial conditions

When the package is dropped from the plane:

Horizontal velocity: vₓ = 81.00 m/s (same as plane)
Vertical velocity: v₀y = 0 m/s (dropped, not thrown)
Height: h = 125 m
Gravity: g = 9.8 m/s²

Step 2 — Calculate the fall time

Use the kinematic equation for vertical motion. Since the package starts with zero vertical velocity:

h = ½gt²
125 = ½(9.8)t²
125 = 4.9t²
t² = 125/4.9 = 25.51
t = √25.51 = 5.05 seconds

Step 3 — Calculate horizontal distance

During the 5.05 seconds of fall time, the package maintains its horizontal velocity of 81.00 m/s:

Distance = velocity × time
Δx = vₓ × t
Δx = 81.00 × 5.05
Δx = 409 meters

Solution: Method 2 — Trajectory Equation Approach

We can also solve this by deriving the parabolic trajectory equation and finding where it intersects the ground.

Step 1 — Set up coordinate system

Place the origin at the drop point, with positive x to the right and positive y downward:

x = vₓt = 81.00t
y = ½gt² = 4.9t²

Step 2 — Eliminate time to get trajectory

From the first equation: t = x/81.00. Substitute into the second equation:

y = 4.9(x/81.00)²
y = 4.9x²/6561
y = 0.000747x²

Step 3 — Find where package hits ground

The package hits the ground when y = 125 meters:

125 = 0.000747x²
x² = 125/0.000747 = 167,336
x = √167,336 = 409 meters
The rescue pilot must drop the package 409 meters horizontally from the target.

Verification

Let's verify by checking our fall time calculation and ensuring the physics makes sense:

Check the fall time

h = ½gt² = ½(9.8)(5.05)² = 4.9 × 25.51 = 125 ✓

Check the horizontal distance

Δx = vₓt = 81.00 × 5.05 = 409 m ✓

Reality check

The package falls for about 5 seconds while traveling at 81 m/s horizontally. This gives roughly 400 meters, which matches our precise calculation. A football field is about 100 meters long, so dropping the package 4 football fields away makes intuitive sense for this speed and height.

Watch Out For These

✗ Confusing the plane's velocity with the package's velocity

The package maintains the plane's horizontal velocity of 81.00 m/s. It does not have an initial vertical velocity of 81.00 m/s - that would be if the pilot threw it downward at 81 m/s.

✗ Using the wrong kinematic equation

Students sometimes try h = v₀t + ½gt² with v₀ = 81 m/s. This treats the 81 m/s as vertical velocity, which is incorrect. The initial vertical velocity is zero when dropping.

✗ Forgetting that gravity affects vertical motion only

Some students think the horizontal distance should account for gravity somehow. In projectile motion (ignoring air resistance), horizontal velocity remains constant throughout the flight.

The Physics Behind This

This problem demonstrates Galilean independence of motion - one of the foundational principles of classical mechanics. Horizontal and vertical motions are completely separate:

General projectile equations:
x(t) = v₀ₓt
y(t) = v₀y·t + ½gt²

For horizontal launch: v₀y = 0
Range formula: R = v₀ₓ√(2h/g)

This principle appears everywhere from ballistics to planetary motion. The same math applies whether you're dropping relief supplies, calculating artillery trajectories, or predicting where a ball will land when thrown from a building.

Beyond the Textbook

This calculation is crucial in several real-world scenarios:

Humanitarian aid drops: Relief organizations use these calculations to ensure supplies land in accessible locations rather than dangerous terrain or water.

Military operations: Paratroop drops and supply missions require precise timing to account for aircraft speed and drop height.

Firefighting: Water and fire retardant drops from aircraft must account for the plane's forward motion to hit the target area effectively.

Space missions: Spacecraft releasing satellites or landing probes use similar projectile motion calculations, though orbital mechanics adds complexity.

What If?

1
Different Height
The same rescue plane traveling at 81.00 m/s is now flying at an altitude of 200 m above the target. How far horizontally from the target should the pilot drop the package?
Step 1 — Find the fall time

Using h = ½gt²: 200 = ½(9.8)t², so t² = 400/9.8 = 40.82 and t = 6.39 seconds

Step 2 — Calculate horizontal distance

Δx = vₓt = 81.00 × 6.39 = 518 meters

Step 3 — Verify

Check: h = ½(9.8)(6.39)² = 200 ✓

Answer: 518 meters from the target

2
Upward Toss
Instead of dropping the package, the pilot gives it an initial upward velocity of 8.0 m/s relative to the plane. The plane still travels at 81.00 m/s horizontally and is 125 m above the target. Where should the pilot release the package?
Step 1 — Set up the problem

Initial conditions: vₓ = 81.0 m/s, v₀y = -8.0 m/s (negative because upward in our coordinate system), h = 125 m

Step 2 — Find total flight time

Using y = v₀y·t + ½gt²: 125 = -8.0t + ½(9.8)t²

4.9t² - 8.0t - 125 = 0

Using quadratic formula: t = (8.0 + √(64 + 2450))/9.8 = 6.13 seconds

Step 3 — Calculate horizontal distance

Δx = 81.0 × 6.13 = 497 meters

Answer: 497 meters from the target

3
Reverse the Unknown
A pilot flying at 125 m altitude drops a package when the plane is exactly 450 m horizontally from the target. For the package to hit the target, what must the plane's horizontal speed be?
Step 1 — Find fall time from height

Using h = ½gt²: 125 = ½(9.8)t², so t = 5.05 seconds

Step 2 — Calculate required horizontal speed

For the package to travel 450 m horizontally in 5.05 seconds:

vₓ = distance/time = 450/5.05 = 89.1 m/s

Step 3 — Verify

Check: distance = 89.1 × 5.05 = 450 ✓

Answer: The plane must travel at 89.1 m/s horizontally

4
Moving Target
The target is on a boat moving at a constant 15 m/s directly away from the plane's flight path. The plane flies at 81.00 m/s and is 125 m high. How far from the target's current position should the pilot drop the package?
Step 1 — Find package fall time

Fall time remains: t = √(2h/g) = √(2×125/9.8) = 5.05 seconds

Step 2 — Calculate target's movement

In 5.05 seconds, the target moves: 15 × 5.05 = 76 meters further away

Step 3 — Calculate package horizontal travel

Package travels: 81.0 × 5.05 = 409 meters horizontally

Step 4 — Find drop distance

Drop distance from current target position = 409 - 76 = 333 meters

Answer: Drop the package 333 meters before the target's current position

Frequently Asked Questions

How do you calculate the fall time for a dropped object?+
Use the kinematic equation h = ½gt² where h is the height, g is gravity (9.8 m/s²), and t is time. Solve for t: t = √(2h/g). In this problem, dropping from 125 m gives t = √(2×125/9.8) = 5.05 seconds.
Why does horizontal velocity stay constant in projectile motion?+
In projectile motion, gravity only acts vertically downward, so there's no force affecting horizontal motion (ignoring air resistance). The package maintains the plane's horizontal velocity of 81 m/s throughout its fall, traveling horizontally while simultaneously falling vertically.
What's the difference between dropping and throwing a projectile?+
Dropping means zero initial vertical velocity - the object only has horizontal velocity from the moving platform. Throwing adds initial vertical velocity (up or down). Here the package is dropped, so it starts with 81 m/s horizontal and 0 m/s vertical velocity.
DN

Dr. Neven Jurkovic

Physics Professor & Educational Content Developer

NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-08-20