Projectile Motion

Learn the models, terms, and applications of projectile motion.

Lesson 2 of 32 parts Practice Flashcards

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Concepts and Models

Objective: Explain the concepts and assumptions used in projectile motion.

In the near-Earth projectile model with negligible drag, horizontal acceleration is zero and vertical acceleration is downward g. The same elapsed time connects both directions. Horizontal displacement is v_x t; vertical displacement follows constant-acceleration equations. At the highest point the vertical velocity is zero, but horizontal velocity and downward acceleration can remain nonzero. The equal-height range formula needs equal launch and landing heights; do not apply it blindly to a launch from a cliff.

Formulas

$Δx=v_x t$
Δy=v_y0t−½gt² with upward positive
maximum rise=v_y0²/(2g)

Worked examples

Try itA projectile has horizontal velocity 5 m/s for 0.5 s with no drag. Find horizontal displacement.Show answer

1. Horizontal acceleration is zero. 2. Δx=v_x t=5×0.5=2.5 m.

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Concepts and Models

Objective: Explain the concepts and assumptions used in projectile motion.

Formulas

$Δx=v_x t$
Δy=v_y0t−½gt² with upward positive
maximum rise=v_y0²/(2g)

Worked examples

Try itA projectile has horizontal velocity 5 m/s for 0.5 s with no drag. Find horizontal displacement.Show answer

1. Horizontal acceleration is zero. 2. Δx=v_x t=5×0.5=2.5 m.

Applications and Worked Problems

Objective: Apply projectile motion to calculations, evidence, and specified situations.

For the application problems, define the system and list the quantities supplied before choosing a relation. Use Δx=v_x t; Δy=v_y0t−½gt² with upward positive; maximum rise=v_y0²/(2g). The relation must match the stated conditions. Convert to compatible units before substitution, preserve intermediate precision, and report the requested quantity with its unit or sign. When comparing alternatives, check the defining condition rather than a superficial word match. The worked examples below show the setup and calculation separately.

Formulas

$Δx=v_x t$
Δy=v_y0t−½gt² with upward positive
maximum rise=v_y0²/(2g)

Worked examples

Try itA projectile has horizontal velocity 5 m/s for 0.5 s with no drag. Find horizontal displacement.Show answer

1. Horizontal acceleration is zero. 2. Δx=v_x t=5×0.5=2.5 m.

Try itA projectile launched horizontally falls for 0.5 s. With g=9.8 m/s², find downward displacement.Show answer

1. Vertical initial velocity is zero. 2. Δy=½gt²=0.5×9.8×0.5²=1.225 m.

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