Topic 1.5: Projectile Motion Problem-Solving

Name: _______________________________ Date: _______________

Before We Start: Yesterday's Recap

Yesterday we learned horizontally-launched projectiles. Without looking back: what's the solving order — vertical first or horizontal first? Why?

Today's New Concepts

AP CED Alignment: Unit 1, Topic 1.5

Angled Launch — Components

• $v_x = v_i\cos\theta$ — stays constant the whole flight
• $v_{y,i} = v_i\sin\theta$ — the initial vertical velocity, undergoes free fall from there

Key Formulas (projectile returns to launch height)

• Time to max height: $t_{up} = \dfrac{v_{y,i}}{g}$
• Total flight time: $2 \times t_{up}$
• Max height: $v_{y,i}^2 = 2g(\text{height}) \rightarrow \text{height} = \dfrac{v_{y,i}^2}{2g}$
• Range: $v_x \times \text{total flight time}$

Worked Example

20 m/s at 30°: $v_x \approx 17.3\text{ m/s}$, $v_{y,i} = 10\text{ m/s}$. $t_{up} \approx 1.02\text{ s}$, total time ≈ 2.04 s. Max height ≈ 5.1 m. Range ≈ 35.3 m.

Why 45° Gives Maximum Range

Range depends on a trade-off: steeper angles give more hang time (more vertical velocity) but less horizontal speed; shallower angles give more horizontal speed but less hang time. 45° is the balance point that maximizes the product of the two — for a fixed launch speed, no other angle produces a longer range.

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Keep This Sheet!

These sheets build into your semester study guide. Keep them in order in a binder or folder — you'll want to flip back through them before quizzes, unit tests, and when AP exam review starts in the spring.