PHY 150 – Project One: Supply Drop Plan
Complete Solutions with Step-by-Step Derivations | 2026 Update
Institution: Southern New Hampshire University (SNHU)
Course: PHY 150 — Introductory Physics
Assessment: Project One — Supply Drop Plan
Questions: 25 Multiple Choice (A–D) with Complete Solutions
Date: July 8, 2026
Academic Year: 2026–2027
Cognitive Level: 30% Recall | 50% Application | 20% Analysis
Standards: SNHU PHY 150 Project Rubric, 2026 University Physics Standards
, SNHU PHY 150 | Project One: Supply Drop Plan | 2026
Base Scenario Parameters
Unless otherwise specified in a question, all 25 questions use the following base scenario parameters:
Aircraft altitude: h = 250 m
Aircraft speed (horizontal): v■■ = 80 m/s
Initial vertical velocity: v■■ = 0 m/s
Gravitational acceleration: g = 9.8 m/s²
Air resistance: None (ideal projectile motion) unless otherwise stated
Pre-calculated Reference Values
Time of flight: t = √(2h/g) = √(500/9.8) = √51.0204 = 7.14 s
Horizontal range: x = v■■ × t = 80 × 7.14 = 571.4 m
Impact vertical velocity: v■ = −g × t = −9.8 × 7.14 = −70.0 m/s
Impact speed: |v| = √(80² + 70²) = √11300 = 106.3 m/s
Impact angle: θ = arctan(70/80) = 41.2° below horizontal
Section 1: Initial Scenario Setup & Vertical Kinematics (Q1–Q6)
Q1. In the context of a supply drop from a level-flying aircraft, which statement correctly describes the initial velocity
vector of the payload at the exact moment of release?
A. The initial velocity is directed straight downward at 9.8 m/s
B. The initial velocity is zero in both the horizontal and vertical directions
C. The initial velocity is purely horizontal, equal to the aircraft’s velocity (80 m/s), with zero vertical component
[CORRECT]
D. The initial velocity has both a horizontal component (80 m/s) and a downward vertical component (9.8 m/s)
Correct Answer: C
Rationale: At the moment of release, the payload shares the aircraft’s horizontal velocity (80 m/s) and has zero initial
vertical velocity (v■■ = 0 m/s) because the aircraft is flying level. The payload does not have a downward component at
release; it begins to accelerate downward only after release due to gravity. Option (A) confuses velocity with acceleration.
Option (B) incorrectly states the initial velocity is zero. Option (D) incorrectly assigns a downward velocity component at
the instant of release.
Q2. An aircraft flying level at an altitude of 250 m releases a supply package. Using g = 9.8 m/s² and assuming no air
resistance, what is the time of flight for the package to reach the ground?
A. 5.05 s
B. 7.14 s [CORRECT]
C. 7.07 s
D. 25.5 s
Correct Answer: B
Rationale: Using the vertical kinematic equation y = y■ + v■■t − ½gt², setting y = 0 (ground level), y■ = 250 m, and
v■■ = 0: 0 = 250 − 4.9t², so t² = 250/4.9 = 51.0204, giving t = 7.14 s. Option (A) results from forgetting the ½ in the
equation (using h = gt² instead of ½gt²): t = √(250/9.8) = 5.05 s. Option (C) results from using g = 10 m/s²: t = √(250/5) =
7.07 s. Option (D) results from incorrectly dividing height by g: t = 250/9.8 = 25.5 s.
Q3. Using the time of flight calculated above (7.14 s), what is the vertical velocity of the package just before impact
with the ground?
A. −35.0 m/s
B. −69.0 m/s
C. −70.0 m/s [CORRECT]
D. −250 m/s
Correct Answer: C
Page 2
Complete Solutions with Step-by-Step Derivations | 2026 Update
Institution: Southern New Hampshire University (SNHU)
Course: PHY 150 — Introductory Physics
Assessment: Project One — Supply Drop Plan
Questions: 25 Multiple Choice (A–D) with Complete Solutions
Date: July 8, 2026
Academic Year: 2026–2027
Cognitive Level: 30% Recall | 50% Application | 20% Analysis
Standards: SNHU PHY 150 Project Rubric, 2026 University Physics Standards
, SNHU PHY 150 | Project One: Supply Drop Plan | 2026
Base Scenario Parameters
Unless otherwise specified in a question, all 25 questions use the following base scenario parameters:
Aircraft altitude: h = 250 m
Aircraft speed (horizontal): v■■ = 80 m/s
Initial vertical velocity: v■■ = 0 m/s
Gravitational acceleration: g = 9.8 m/s²
Air resistance: None (ideal projectile motion) unless otherwise stated
Pre-calculated Reference Values
Time of flight: t = √(2h/g) = √(500/9.8) = √51.0204 = 7.14 s
Horizontal range: x = v■■ × t = 80 × 7.14 = 571.4 m
Impact vertical velocity: v■ = −g × t = −9.8 × 7.14 = −70.0 m/s
Impact speed: |v| = √(80² + 70²) = √11300 = 106.3 m/s
Impact angle: θ = arctan(70/80) = 41.2° below horizontal
Section 1: Initial Scenario Setup & Vertical Kinematics (Q1–Q6)
Q1. In the context of a supply drop from a level-flying aircraft, which statement correctly describes the initial velocity
vector of the payload at the exact moment of release?
A. The initial velocity is directed straight downward at 9.8 m/s
B. The initial velocity is zero in both the horizontal and vertical directions
C. The initial velocity is purely horizontal, equal to the aircraft’s velocity (80 m/s), with zero vertical component
[CORRECT]
D. The initial velocity has both a horizontal component (80 m/s) and a downward vertical component (9.8 m/s)
Correct Answer: C
Rationale: At the moment of release, the payload shares the aircraft’s horizontal velocity (80 m/s) and has zero initial
vertical velocity (v■■ = 0 m/s) because the aircraft is flying level. The payload does not have a downward component at
release; it begins to accelerate downward only after release due to gravity. Option (A) confuses velocity with acceleration.
Option (B) incorrectly states the initial velocity is zero. Option (D) incorrectly assigns a downward velocity component at
the instant of release.
Q2. An aircraft flying level at an altitude of 250 m releases a supply package. Using g = 9.8 m/s² and assuming no air
resistance, what is the time of flight for the package to reach the ground?
A. 5.05 s
B. 7.14 s [CORRECT]
C. 7.07 s
D. 25.5 s
Correct Answer: B
Rationale: Using the vertical kinematic equation y = y■ + v■■t − ½gt², setting y = 0 (ground level), y■ = 250 m, and
v■■ = 0: 0 = 250 − 4.9t², so t² = 250/4.9 = 51.0204, giving t = 7.14 s. Option (A) results from forgetting the ½ in the
equation (using h = gt² instead of ½gt²): t = √(250/9.8) = 5.05 s. Option (C) results from using g = 10 m/s²: t = √(250/5) =
7.07 s. Option (D) results from incorrectly dividing height by g: t = 250/9.8 = 25.5 s.
Q3. Using the time of flight calculated above (7.14 s), what is the vertical velocity of the package just before impact
with the ground?
A. −35.0 m/s
B. −69.0 m/s
C. −70.0 m/s [CORRECT]
D. −250 m/s
Correct Answer: C
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