Two balls thrown vertically — one up at 20 m/s, one down from 40m height — when do they meet

Question

Ball A is thrown vertically upward from the ground with a speed of 20 m/s. At the same instant, Ball B is dropped (released from rest) from a height of 40 m. When and where do the two balls meet? (Take $g = 10$ m/s²)

Solution — Step by Step

Set up a coordinate system

We take upward as positive, and the ground (where Ball A starts) as origin ( $y = 0$ ).

Ball A: starts at $y = 0$ , initial velocity $u_A = +20$ m/s (upward), acceleration $a = -10$ m/s² (gravity, downward).

Ball B: starts at $y = 40$ m, initial velocity $u_B = 0$ (dropped, not thrown), acceleration $a = -10$ m/s² (gravity, downward).

Both balls are launched at the same instant $t = 0$ .

Write position equations for both balls

Using $y = y_0 + ut + \frac{1}{2}at^2$ :

Ball A: $y_A = 0 + 20t + \frac{1}{2}(-10)t^2 = 20t - 5t^2$

Ball B: $y_B = 40 + 0 \cdot t + \frac{1}{2}(-10)t^2 = 40 - 5t^2$

Set positions equal to find when they meet

The balls meet when $y_A = y_B$ :

$20t - 5t^2 = 40 - 5t^2$

Notice that the $-5t^2$ terms cancel! This is elegant — it means the meeting time is independent of $g$ entirely.

$20t = 40$

$\boxed{t = 2 \text{ s}}$

The balls meet 2 seconds after launch.

Find where they meet

Substitute $t = 2$ into either position equation:

$y_A = 20(2) - 5(2)^2 = 40 - 20 = 20 \text{ m}$

Verify with Ball B: $y_B = 40 - 5(4) = 40 - 20 = 20$ m. ✓

$\boxed{\text{They meet at a height of 20 m above the ground}}$

Why This Works

The cancellation of $-5t^2$ terms is the key insight. Both balls experience the same gravitational acceleration, so their downward displacement due to gravity is identical ( $\frac{1}{2}gt^2$ downward for both). Gravity effectively "disappears" from the problem — we only need to consider their initial separation (40 m) and the closing speed.

Think of it this way: in the reference frame of Ball B (which is in free fall), gravity is cancelled out. Ball A appears to move at a constant velocity of 20 m/s toward Ball B (since both experience the same $g$ , the relative acceleration is zero). Time to close a gap of 40 m at 20 m/s: $t = 40/20 = 2$ s. Exactly what we found.

This is a beautiful application of the principle of relative motion: when two objects have the same acceleration, you can analyse their relative motion as if there were no acceleration at all.

Alternative Method

Using relative motion directly:

Relative velocity of A with respect to B = $v_A - v_B = 20 - 0 = 20$ m/s (upward at $t=0$ ).

Since both have the same acceleration $(-g)$ , the relative acceleration = $(-g) - (-g) = 0$ .

So A approaches B at a constant velocity of 20 m/s. Initial separation = 40 m.

Time to meet: $t = \frac{40}{20} = 2$ s.

Height at meeting: use Ball B: $y = 40 - \frac{1}{2}(10)(2^2) = 40 - 20 = 20$ m.

This is faster and conceptually cleaner.

Common Mistake

⚠️ Common Mistake

A common error is forgetting to check that the balls actually meet before Ball A reaches its maximum height and before Ball B hits the ground. Ball A reaches maximum height at $t = u/g = 20/10 = 2$ s — so they meet exactly when Ball A is at its peak! Ball B hits the ground when $40 - 5t^2 = 0 \Rightarrow t = 2\sqrt{2} \approx 2.83$ s, so they do meet before that. Always verify that the time of meeting is physically valid.

💡 Expert Tip

The fact that $g$ cancels is not a coincidence — it will always happen when two objects are projected simultaneously under gravity. The meeting time depends only on the initial separation and the initial relative velocity. This is a useful shortcut for MCQs.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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