Question

1. what is the distance fallen for a freely falling object 1 second after being dropped from a rest position? what is the distance for a 4 - second drop?

Explanation:

Step1: Identify the formula

The formula for the distance fallen by a freely - falling object is $d = v_0t+\frac{1}{2}gt^2$. Since the object is dropped from rest, $v_0 = 0$. So the formula simplifies to $d=\frac{1}{2}gt^2$, where $g = 9.8\ m/s^2$ (acceleration due to gravity near the Earth's surface).

Step2: Calculate distance for $t = 1\ s$

Substitute $t = 1\ s$ and $g=9.8\ m/s^2$ into $d=\frac{1}{2}gt^2$.
$d_1=\frac{1}{2}\times9.8\times1^2=4.9\ m$

Step3: Calculate distance for $t = 4\ s$

Substitute $t = 4\ s$ and $g = 9.8\ m/s^2$ into $d=\frac{1}{2}gt^2$.
$d_2=\frac{1}{2}\times9.8\times4^2=\frac{1}{2}\times9.8\times16 = 78.4\ m$

Answer:

The distance fallen in 1 second is $4.9\ m$ and the distance fallen in 4 seconds is $78.4\ m$.