CLASSICAL-MECHANICS · Interactive Practice | Unit 4 · Video 5

Aim Straight at the Falling Bucket

IKey Formulas

The thrower releases the ball at height $h_2$ a horizontal distance $s$ from the ladder; the pail is released from rest at height $h_1$ .

Formula	Name	What it tells you
$\tan\theta_0 = \dfrac{h_1 - h_2}{s}$	Aiming angle	Aim straight at the pail's start — independent of $v_0$
$t_a = \dfrac{1}{v_0}\sqrt{s^2 + (h_1 - h_2)^2}$	Collision time	Straight-line distance divided by launch speed
$y_1(t_a) = h_1 - \dfrac{g\left(s^2 + (h_1 - h_2)^2\right)}{2 v_0^2}$	Collision height	Pail's start, minus how far it has fallen
$\sin^2\theta_0 + \cos^2\theta_0 = 1$	Pythagorean identity	The trick that collapses the squared equations

IIVisualization 1 — Aiming Angle Ignores Speed

The aiming angle $\theta_0 = \arctan\dfrac{h_1 - h_2}{s}$ is fixed by the geometry — does the launch speed change it?

Pail height h₁ 8.0 Release height h₂ 2.0 Distance s 6.0 Launch speed v₀ 15

IIIVisualization 2 — Gravity Cancels in the Falling Frame

Watch from a frame falling at $g$ : gravity vanishes, the pail hangs still, and the ball flies dead straight at it.

t / tₐ 0.60

💡 Both ball and pail fall the same $\tfrac{1}{2} g t^2$ , so that term cancels in the height equation — which is why the aim is purely geometric.

IVVisualization 3 — Collision Time vs. Launch Speed

With the geometry fixed, what shape does the collision time $t_a$ trace as the launch speed grows?

Launch speed v₀ 15

💡 The numerator $\sqrt{s^2 + (h_1 - h_2)^2}$ depends only on the geometry, so $t_a \propto 1/v_0$ : throw twice as hard, meet in half the time.

VQuiz Questions

Question 1

The thrower releases the ball at height $h_2 = 2$ m, a horizontal distance $s = 6$ m from a ladder. The pail is released from rest at height $h_1 = 8$ m. At what angle above the horizontal should the ball be aimed?

Use $\tan\theta_0 = \dfrac{h_1 - h_2}{s}$ .

✅ Correct! With rise = run = 6, the tangent is 1, so θ₀ = 45°.

❌ Not quite. Compute tan θ₀ = (h₁ − h₂)/s = (8 − 2)/6 = 1, then take the arctangent.

Show solution

Solution:

The aiming angle depends only on the geometry:

$\tan\theta_0 = \frac{h_1 - h_2}{s} = \frac{8 - 2}{6} = \frac{6}{6} = 1$

So $\theta_0 = \arctan(1) = 45^\circ$ . The launch speed $v_0$ never enters — you aim straight at the pail's starting position regardless of how hard you throw.

Question 2

True or False: If the thrower doubles the launch speed $v_0$ (keeping $h_1$ , $h_2$ , and $s$ the same), the aiming angle $\theta_0$ needed to hit the pail must be increased.

✅ Correct! The angle depends only on (h₁ − h₂)/s — v₀ never appears.

❌ Not quite. The aiming angle is independent of v₀; gravity's drop cancels for both objects.

Show solution

Solution:

False. The aiming angle is

$\tan\theta_0 = \frac{h_1 - h_2}{s},$

which contains no $v_0$ . In the height condition, the $-\tfrac{1}{2}g t_a^2$ term appears identically for both the ball and the pail, so it cancels — gravity drops them both by the same amount. The angle is set purely by the geometry. Doubling $v_0$ only makes the collision happen sooner (and higher), not at a different angle.

Question 3

Using the same setup ( $h_1 = 8$ m, $h_2 = 2$ m, $s = 6$ m), the ball is thrown with launch speed $v_0 = 12$ m/s. How long after release do the ball and pail collide?

Use $t_a = \dfrac{1}{v_0}\sqrt{s^2 + (h_1 - h_2)^2}$ .

✅ Correct! The diagonal distance is 6√2 ≈ 8.49 m, and 8.49 / 12 ≈ 0.71 s.

❌ Not quite. Use the full straight-line distance √(s² + (h₁ − h₂)²), not just s, then divide by v₀.

Show solution

Solution:

The straight-line distance from the release point to the pail's start is

$\sqrt{s^2 + (h_1 - h_2)^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \approx 8.49\ \text{m}.$

Dividing by the launch speed:

$t_a = \frac{1}{v_0}\sqrt{s^2 + (h_1 - h_2)^2} = \frac{8.49}{12} \approx 0.71\ \text{s}.$

A common mistake is forgetting the vertical leg and using only $s/v_0 = 6/12 = 0.50$ s — but the ball must cover the full diagonal distance.

Question 4

In the algebra that gives the collision time, we square the two equations $v_0 \sin\theta_0\, t_a = h_1 - h_2$ and $v_0 \cos\theta_0\, t_a = s$ , then add them. Which identity makes the right-hand approach collapse to $v_0^2 t_a^2 = s^2 + (h_1 - h_2)^2$ ?

✅ Correct! The Pythagorean identity turns the sin² + cos² factor into 1.

❌ Not quite. The factored term is sin²θ₀ + cos²θ₀, which equals 1 by the Pythagorean identity.

Show solution

Solution:

Squaring and adding the two equations gives

$v_0^2 t_a^2 \sin^2\theta_0 + v_0^2 t_a^2 \cos^2\theta_0 = (h_1 - h_2)^2 + s^2.$

Factoring the left side:

$v_0^2 t_a^2 \left(\sin^2\theta_0 + \cos^2\theta_0\right) = s^2 + (h_1 - h_2)^2.$

The Pythagorean identity $\sin^2\theta_0 + \cos^2\theta_0 = 1$ collapses the parenthesis to $1$ , leaving $v_0^2 t_a^2 = s^2 + (h_1 - h_2)^2$ . Taking the square root gives $v_0 t_a = \sqrt{s^2 + (h_1 - h_2)^2}$ .

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