Final answer:
The freely falling body would cover a distance of 19.6 meters in the first 2 seconds of its fall. To calculate the distance covered by a freely falling body in 2 seconds from rest, we can use the kinematic equation for uniformly accelerated motion with Earth's gravity as the acceleration.
Step-by-step explanation:
The student is asking about the distance covered by a freely falling body in the first 2 seconds of its fall, with zero initial velocity. This situation relates to the physics of free fall under the influence of gravity.
Using the formula for the distance covered (s) under uniform acceleration (a), which for Earth's gravity is approximately 9.80 m/s², we have:
s = Vi·t + (1/2)·a·t²
Where:
- Vi = initial velocity (0 m/s since it starts from rest)
- t = time (2 seconds)
- a = acceleration due to gravity (9.80 m/s²)
Plugging the values into the equation, we get:
s = 0·(2) + (1/2)·(9.80)·(2)²
Therefore, s = (1/2)·(9.80)·(4)
= 19.6 meters
The distance covered in this time period is found to be 19.6 meters.