Solution 5: Colliding Balls

Solution 5

Time Sequence Logic: Let the total height of the cliff be $H$. Let $T$ be the time it takes for an object to free-fall from top to ground ($T = \sqrt{2H/g}$).

Ball 1: Drops at $t=0$, hits ground at $t=T$, rebounds elastically, and moves up. At the moment of collision, it has been moving up for time $\tau$. Total time elapsed $t_{total} = T + \tau$.
Ball 2: Drops at $t = \Delta t = 2$ s. It has been falling for time $(t_{total} – 2)$.

The “collision” means Ball 1 (moving up) and Ball 2 (moving down) meet at the same height $h=55$ m.

1. Analyzing Ball 2 (The Falling Ball)

Ball 2 is dropped from the top. It reaches height $h=55$ m.

$$ y = H – \frac{1}{2}gt^2 \implies 55 = H – \frac{1}{2}g(t_{fall,2})^2 $$

Rearranging for the distance fallen:

$$ \frac{1}{2}g(t_{fall,2})^2 = H – 55 \quad \text{— (Eq 1)} $$

2. Analyzing Ball 1 (The Rising Ball)

Ball 1 fell for time $T$, hit the ground, and rebounded with speed $u = gT$ (elastic collision conserves speed). It then rose to height $h=55$ m in time $\tau$.

$$ h = u\tau – \frac{1}{2}g\tau^2 \implies 55 = (gT)\tau – \frac{1}{2}g\tau^2 $$

Let’s substitute $H = \frac{1}{2}gT^2$ into Eq 1:

$$ \frac{1}{2}g(t_{fall,2})^2 = \frac{1}{2}gT^2 – 55 \implies 55 = \frac{1}{2}g(T^2 – t_{fall,2}^2) $$

We equate the two expressions for 55:

$$ gT\tau – \frac{1}{2}g\tau^2 = \frac{1}{2}g(T^2 – t_{fall,2}^2) $$

Divide by $g/2$:

$$ 2T\tau – \tau^2 = T^2 – t_{fall,2}^2 $$

3. Solving for Time

We need to relate the times. Let the collision happen at absolute time $t_c$.

Ball 1 hit ground at $T$. So $\tau = t_c – T$.
Ball 2 was dropped at $t=2$. So $t_{fall,2} = t_c – 2$.

Notice the relationship: $t_{fall,2} = (T + \tau) – 2$. Let’s substitute $t_{fall,2}$ into the simplified equation:

$$ 2T\tau – \tau^2 = T^2 – (T + \tau – 2)^2 $$ $$ 2T\tau – \tau^2 = T^2 – [ (T+\tau) – 2 ]^2 $$ $$ 2T\tau – \tau^2 = T^2 – [ (T+\tau)^2 – 4(T+\tau) + 4 ] $$ $$ 2T\tau – \tau^2 = T^2 – [ T^2 + 2T\tau + \tau^2 – 4T – 4\tau + 4 ] $$ $$ 2T\tau – \tau^2 = -2T\tau – \tau^2 + 4T + 4\tau – 4 $$

Simplifying:

$$ 4T\tau = 4T + 4\tau – 4 $$ $$ T\tau = T + \tau – 1 \implies T\tau – T – \tau + 1 = 0 $$ $$ (T-1)(\tau-1) = 0 $$

This gives two possibilities: $T=1$ or $\tau=1$.

If $T=1$: Max height $H = 5(1)^2 = 5$ m. But collision is at 55 m. Impossible.
Therefore, $\tau = 1$ s.

4. Calculating Height

Now we use the collision height equation for Ball 1 with $\tau = 1$:

$$ 55 = gT(1) – \frac{1}{2}g(1)^2 $$ $$ 55 = 10T – 5 \implies 60 = 10T \implies T = 6 \text{ s} $$

Finally, the height of the cliff is:

$$ H = \frac{1}{2}gT^2 = \frac{1}{2}(10)(6^2) = 5(36) = 180 \text{ m} $$

Physics Problem Solution

ALITER

Step 1: Analyze the Timeline

Let’s define the key points in the motion:

A: Top of the cliff.
B: Ground.
C: Collision point (55m above ground).

Ball 1 is dropped first. Ball 2 is dropped 2 seconds later.

At the moment of collision (at point C):

Ball 2 is on its way down (A → C).
Ball 1 has gone down, hit the ground, and is on its way up (A → B → C).

Since both balls fall from rest from A, the time taken to travel A → C is identical for both. Let’s call this time $ t_{AC} $.

Because Ball 1 was released 2 seconds earlier, it has had 2 extra seconds of travel time. These extra 2 seconds account for the trip from C down to the ground (B) and back up to C.

t_{C \to B} + t_{B \to C} = \Delta t = 2 \text{ s}

Due to symmetry (elastic collision, no air resistance), time down equals time up:

t_{C \to B} = 1 \text{ second}

Step 2: Calculate Velocity at Point C

Now consider the motion of Ball 1 falling from C to B.

Distance $ s = h = 55 \text{ m} $
Time $ t = 1 \text{ s} $
Acceleration $ a = g = 10 \text{ m/s}^2 $
Initial Velocity at C $ = v_C $

Using the kinematic equation $ s = ut + \frac{1}{2}at^2 $:

55 = v_C(1) + \frac{1}{2}(10)(1)^2 \] \[ 55 = v_C + 5 \] \[ v_C = 50 \text{ m/s}

Step 3: Calculate Total Height

Now we calculate the total fall. First, find how long it takes to reach speed $ v_C = 50 \text{ m/s} $ from rest (falling from top A to C).

Using $ v = u + at $:

50 = 0 + 10(t_{AC}) \] \[ t_{AC} = 5 \text{ s}

The total time to fall from the top of the cliff to the ground (A → B) is:

T_{total} = t_{AC} + t_{CB} = 5 + 1 = 6 \text{ s}

Finally, calculate the total height $ H $ of the cliff:

H = \frac{1}{2} g T_{total}^2 \] \[ H = \frac{1}{2} (10) (6)^2 \] \[ H = 5 \times 36 \] \[ H = 180 \text{ m}

Answer

The height of the top of the cliff is 180 meters.