PHY 150
PROJECT ONE: SUPPLY DROP PLAN
2026 Update | With Complete Solutions
Institution: Southern New Hampshire University
Course: PHY 150 - Introductory Physics
Document Type: Project One Exam with Solutions
Total Questions: 25 (Multiple Choice)
Total Points: 100 (4 points per question)
Topics: 2D Kinematics, Projectile Motion, Vector Analysis, Trajectory Calculations
Date: July 8, 2026
This exam assesses mastery of 2D kinematic principles applied to supply drop trajectory
calculations, including vector resolution of initial velocities, independent horizontal and
vertical motion analysis, recalculation under variable conditions, and modern
autonomous delivery system concepts.
, PHY 150 Project One | Supply Drop Plan | 2026 Update
Scenario Context
A humanitarian aid aircraft is flying horizontally at a constant speed of 80.0 m/s at an altitude of 500 m above flat
terrain. The mission is to deliver a supply payload to a designated ground target. At the moment of release, the
payload inherits the aircraft's horizontal velocity and has an initial vertical velocity of zero. Use g = 9.80 m/s²
throughout. Unless otherwise stated, ignore air resistance. All questions reference this base scenario unless a
variation is explicitly introduced.
Key Pre-Computed Values (Base Scenario):
Parameter Value Equation
Time of flight (t) 10.10 s t = √(2h/g) = √(2×500/9.80)
Horizontal range (x) 808.2 m x = v■x × t = 80.0 × 10.10
Vertical impact velocity (v_y) 98.98 m/s v_y = g × t = 9.80 × 10.10
Impact speed (|v|) 127.3 m/s |v| = √(v_x² + v_y²)
Impact angle (θ) 51.0° below horizontal θ = arctan(v_y/v_x)
Section 1: Initial Scenario Setup & Vertical Kinematics
This section establishes the foundational vertical kinematics of the supply drop. Questions assess understanding of
initial conditions, correct equation setup, time-of-flight derivation, and the relationship between altitude and fall
duration.
Q1: At the exact moment a supply payload is released from a horizontally flying aircraft at 80.0 m/s, what is
the initial vertical velocity (v0y) of the payload?
A. 0 m/s, because the payload has no vertical motion component at the instant of release
B. 80.0 m/s, because it inherits the full aircraft velocity
C. 9.80 m/s² downward, because gravity acts immediately
D. Equal to the aircraft's vertical climb rate, which is unknown
Correct Answer: A
Rationale: The payload is moving horizontally with the aircraft; at the instant of release, the vertical component of
velocity is zero. The horizontal velocity (80.0 m/s) is preserved, but v0y = 0. Choice B confuses horizontal with vertical
velocity. Choice C incorrectly provides the gravitational acceleration (not a velocity). Choice D introduces an
unnecessary assumption about vertical climb rate when the problem specifies level flight.
Q2: Which kinematic equation correctly represents the vertical displacement to solve for time of flight, taking
upward as the positive direction, for a payload dropped from y0 = 500 m?
A. y = 0.5 × g × t²
B. 0 = 500 − 0.5 × (9.80) × t²
C. y = v0y × t + 0.5 × g × t²
D. vy = v0y + g × t
Correct Answer: B
Rationale: Starting from y = y0 + v0yt − 0.5gt², with y0=500 m, v0y=0, and final y=0: 0 = 500 − 0.5(9.80)t². Choice A
omits the initial height. Choice C uses positive g (wrong sign for downward acceleration). Choice D is a velocity
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PROJECT ONE: SUPPLY DROP PLAN
2026 Update | With Complete Solutions
Institution: Southern New Hampshire University
Course: PHY 150 - Introductory Physics
Document Type: Project One Exam with Solutions
Total Questions: 25 (Multiple Choice)
Total Points: 100 (4 points per question)
Topics: 2D Kinematics, Projectile Motion, Vector Analysis, Trajectory Calculations
Date: July 8, 2026
This exam assesses mastery of 2D kinematic principles applied to supply drop trajectory
calculations, including vector resolution of initial velocities, independent horizontal and
vertical motion analysis, recalculation under variable conditions, and modern
autonomous delivery system concepts.
, PHY 150 Project One | Supply Drop Plan | 2026 Update
Scenario Context
A humanitarian aid aircraft is flying horizontally at a constant speed of 80.0 m/s at an altitude of 500 m above flat
terrain. The mission is to deliver a supply payload to a designated ground target. At the moment of release, the
payload inherits the aircraft's horizontal velocity and has an initial vertical velocity of zero. Use g = 9.80 m/s²
throughout. Unless otherwise stated, ignore air resistance. All questions reference this base scenario unless a
variation is explicitly introduced.
Key Pre-Computed Values (Base Scenario):
Parameter Value Equation
Time of flight (t) 10.10 s t = √(2h/g) = √(2×500/9.80)
Horizontal range (x) 808.2 m x = v■x × t = 80.0 × 10.10
Vertical impact velocity (v_y) 98.98 m/s v_y = g × t = 9.80 × 10.10
Impact speed (|v|) 127.3 m/s |v| = √(v_x² + v_y²)
Impact angle (θ) 51.0° below horizontal θ = arctan(v_y/v_x)
Section 1: Initial Scenario Setup & Vertical Kinematics
This section establishes the foundational vertical kinematics of the supply drop. Questions assess understanding of
initial conditions, correct equation setup, time-of-flight derivation, and the relationship between altitude and fall
duration.
Q1: At the exact moment a supply payload is released from a horizontally flying aircraft at 80.0 m/s, what is
the initial vertical velocity (v0y) of the payload?
A. 0 m/s, because the payload has no vertical motion component at the instant of release
B. 80.0 m/s, because it inherits the full aircraft velocity
C. 9.80 m/s² downward, because gravity acts immediately
D. Equal to the aircraft's vertical climb rate, which is unknown
Correct Answer: A
Rationale: The payload is moving horizontally with the aircraft; at the instant of release, the vertical component of
velocity is zero. The horizontal velocity (80.0 m/s) is preserved, but v0y = 0. Choice B confuses horizontal with vertical
velocity. Choice C incorrectly provides the gravitational acceleration (not a velocity). Choice D introduces an
unnecessary assumption about vertical climb rate when the problem specifies level flight.
Q2: Which kinematic equation correctly represents the vertical displacement to solve for time of flight, taking
upward as the positive direction, for a payload dropped from y0 = 500 m?
A. y = 0.5 × g × t²
B. 0 = 500 − 0.5 × (9.80) × t²
C. y = v0y × t + 0.5 × g × t²
D. vy = v0y + g × t
Correct Answer: B
Rationale: Starting from y = y0 + v0yt − 0.5gt², with y0=500 m, v0y=0, and final y=0: 0 = 500 − 0.5(9.80)t². Choice A
omits the initial height. Choice C uses positive g (wrong sign for downward acceleration). Choice D is a velocity
Page 2
