An object is released from rest at t = 0. How far does it fall during the first second after release?Neglect air friction and express your answer in meters.

1. Break Down the Problem

We need to determine the distance an object falls in the first second after being released from rest. The relevant factors include:

• Initial velocity ($v_0$) = 0 m/s (since the object is released from rest)
• Acceleration due to gravity ($g$) = $9.81$ m/s²
• Time ($t$) = 1 second

2. Relevant Concepts

To find the distance fallen, we can use the formula for distance under constant acceleration: $d = v_0 t + \frac{1}{2} g t^2$ where:

• $d$ = distance fallen
• $v_0$ = initial velocity
• $g$ = acceleration due to gravity
• $t$ = time elapsed

3. Analysis and Detail

Now substituting the known values into the formula:

1. Initial velocity $v_0 = 0$ m/s
2. Acceleration due to gravity $g = 9.81$ m/s²
3. Time $t = 1$ second

Thus, the equation simplifies to: $d = 0 \cdot 1 + \frac{1}{2} \cdot 9.81 \cdot (1^2)$ Calculating further: $d = 0 + \frac{1}{2} \cdot 9.81 \cdot 1$ $d = \frac{9.81}{2} = 4.905 \text{ m}$

4. Verify and Summarize

The distance fallen in the first second is $4.905$ meters. This calculation is verified by following the standard kinematics equations, ensuring all values were substituted correctly.

Final Answer

The object falls approximately 4.905 meters during the first second after release.