4.6 — Worked problems — Chapter 4 toolkit in action

Seven worked problems tagged with the relevant §4.x section.

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Problem 1 (§4.3) — throwing a ball to the ceiling

Ball launched at v_0 = 25 m/s, \theta = 55°. Peak height and range?

Solution:

• v_{0x} = 25\cos 55° \approx 14.34\ m/s
• v_{0y} = 25\sin 55° \approx 20.48\ m/s
• y_\text{max} = v_{0y}^2/(2g) \approx 21.4\ m
• T = 2v_{0y}/g \approx 4.18\ s
• R = v_{0x} T \approx 60.0\ m

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Problem 2 (§4.3) — launched from a cliff

From a 50 m cliff, throw a stone horizontally at v_0 = 20 m/s. Where does it land? Impact speed?

Solution:

• t = \sqrt{100/9.8} \approx 3.19\ s
• x = 20 \cdot 3.19 = 63.9\ m (from cliff base)
• v_y = -31.3\ m/s
• Impact: \sqrt{20^2 + 31.3^2} \approx 37.1\ m/s

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Problem 3 (§4.4) — spinning disc

A pod on a sphere of radius 10 m moving at 50 m/s. Centripetal acceleration and period?

Solution:

• a_c = 50^2/10 = 250\ m/s^2 \approx 25.5g
• T = 2\pi r/v \approx 1.26\ s

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Problem 4 (§4.4) — satellite in orbit

A satellite in a circular orbit at r = 7000 km. Orbital speed? (Assume g = 8.14 m/s^2 at that radius.)

Solution: Centripetal acceleration = gravity:

• v = \sqrt{gr} = \sqrt{8.14 \cdot 7 \times 10^6} \approx 7550\ m/s

Period: T \approx 5827\ s \approx 97\ min

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Problem 5 (§4.5) — river crossing

River of width 100 m, flowing 2 m/s. Boat does 4 m/s in water.

a) If boat aims straight across, where does it land and how long does it take? b) To land directly across (no drift), what angle?

Solution:

a) Crossing time: t = 100/4 = 25\ s. Drift: x = 2 \cdot 25 = 50\ m downstream.

b) Must head \sin\theta = 2/4 = 0.5 \Rightarrow \theta = 30° upstream. Ground speed: 4\cos 30° \approx 3.46\ m/s. Time: ~28.9\ s (longer than a).

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Problem 6 (§4.3 + §4.4) — projectile with spin (curveball)

Baseball at v_0 = 40 m/s, \theta = 30°. Plus a lateral acceleration of 1.5 m/s^2 perpendicular to velocity (Magnus effect from spin, to the right). Compare with standard projectile — approximate lateral deflection?

Simplified solution:

• Standard range: R_0 = v_0^2 \sin(2\theta)/g \approx 141.4\ m
• Flight time: T \approx 4.08\ s
• Lateral deflection: \frac{1}{2}(1.5)(4.08)^2 \approx 12.5\ m

The ball drifts ~12 m to the right. That's the basis of the curveball.

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Problem 7 (§4.2 + §4.3) — particle with time-varying acceleration

At t=0, particle is at origin with \vec v_0 = 3\hat i\ m/s. Acceleration \vec a(t) = 2t\hat j\ m/s^2. Position \vec r(t)?

Solution:

\[ \vec v(t) = 3\hat i + t^2\hat j \]

\[ \vec r(t) = 3t\hat i + \frac{t^3}{3}\hat j \]

At t = 3: \vec r = 9\hat i + 9\hat j, \vec v = 3\hat i + 9\hat j.

Insight: even for time-dependent acceleration, the §2.6 integral tools work — just vector-valued.

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Final strategy

1. Choose a frame and axes that simplify the problem
2. Decompose vectors into components
3. Solve independent x, y components — this is 90% of 2D problems
4. Recombine the answer as magnitude/angle or vector
5. Sanity check — signs, units, magnitude

Preview of §4.7

Twenty practice problems (no step-by-step solutions) with final answers.

📚 See also: Halliday Vol 1, Ch 4 — Worked Examples.

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