Calculate the vertical distance an object dropped from rest covers in 12 seconds of free fall. (The unit of measure for distance is the meter, written as m.)

s = 1/2 at^2

= 1/2 * 9.8 * 12^2
= 705.6

Wait. Won't you have to use the equation yf=yi+viy+1/2ayt2?

To calculate the vertical distance an object dropped from rest covers in 12 seconds of free fall, we can use the formula:

distance = 1/2 * acceleration * time^2

The acceleration due to gravity is approximately 9.8 m/s^2 (rounded to one decimal place).

Plugging in the values:

distance = 1/2 * 9.8 m/s^2 * (12 s)^2

Solving this equation step-by-step:

1. Calculate the time squared:
(12 s)^2 = 144 s^2

2. Multiply the acceleration and time squared:
1/2 * 9.8 m/s^2 * 144 s^2 = 70.56 m^2/s^2

3. Simplify the units:
70.56 m^2/s^2 = 70.56 m

Therefore, an object dropped from rest covers a vertical distance of 70.56 meters in 12 seconds of free fall.

To calculate the vertical distance an object dropped from rest covers in 12 seconds of free fall, we can use the equation for distance covered during free fall:

distance = 0.5 * acceleration * time^2

In this case, the object is dropped from rest, so its initial velocity (u) is 0 m/s. The acceleration due to gravity (g) is approximately 9.8 m/s^2.

Substituting the given values into the equation:

distance = 0.5 * 9.8 * (12^2)
= 0.5 * 9.8 * 144
= 0.5 * 1411.2
= 705.6 m

Therefore, the vertical distance covered by the object in 12 seconds of free fall is 705.6 meters.