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Free Fall Motion

CSCA Free Fall Motion study guide organized around the publicly available CSCA syllabus. Practice Physics questions on aicsca.com.

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Topic: Free Fall Motion

#### 1. Core Definition & Conditions

Free fall is a specific type of **Uniformly Accelerated Linear Motion** with zero initial velocity, where the acceleration is the constant gravitational acceleration $g$.

* **Ideal Conditions**:

1. **Initial Velocity is Zero** ($v_0 = 0$).

2. **Gravity Only**: Air resistance is neglected. If air resistance is significant, it is NOT free fall.

* **Acceleration due to Gravity ($g$)**:

* **Direction**: Vertically Downward.

* **Magnitude**: Usually $g \approx 9.8 m/s^2$ or simplified to $10 m/s^2$ (Check exam instructions).

#### 2. Core Formulas

Free fall is a special case where $v_0=0, a=g$. We usually define **downward as positive**.

| Quantity | Formula | Note |

| :--- | :--- | :--- |

| **Velocity** | $$v = gt$$ | Velocity is proportional to time |

| **Displacement (Height)** | $$h = \frac{1}{2}gt^2$$ | Height is proportional to time squared |

| **Velocity-Height** | $$v^2 = 2gh$$ | Best choice when time $t$ is unknown |

**Important Inferences (Shortcuts)**:

* **Time to Fall**: $t = \sqrt{\frac{2h}{g}}$ (Depends only on height).

* **Impact Velocity**: $v = \sqrt{2gh}$ (Independent of mass, depends only on height).

#### 3. Proportional Patterns

For uniformly accelerated motion starting from rest, these ratios are useful for quick solving:

* **Velocity at end of $1T, 2T, 3T...$**:

$$v_1 : v_2 : v_3 = 1 : 2 : 3$$

* **Total Displacement in $1T, 2T, 3T...$**:

$$h_1 : h_2 : h_3 = 1^2 : 2^2 : 3^2 = 1 : 4 : 9$$

* **Displacement during 1st, 2nd, 3rd interval...**:

$$\Delta h_I : \Delta h_{II} : \Delta h_{III} = 1 : 3 : 5$$

#### 4. Typical Example: Reaction Time Test

**Problem**: Student A holds the top of a $30cm$ ruler. Student B is ready at the $0$ mark at the bottom. A drops it, and B catches it at the $20cm$ mark. Find B's reaction time. ($g=10 m/s^2$)

**Solution**:

The ruler undergoes free fall. Distance $h = 0.2 m$.

From $h = \frac{1}{2}gt^2$:

$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.2}{10}} = \sqrt{0.04} = 0.2s$$

**Common Mistakes**:

* "Distance in the last second": Do not substitute $t=1$. You must set total time as $t$ and solve the equation.

* **Galileo's Experiment**: Heavy and light objects fall at the same rate (in vacuum). If air resistance exists, heavier objects often fall faster (relative effect of resistance is smaller).