Question

A ball falls from rest in the absence of air resistance. The position of the center of the ball is determined at one-second intervals from the instant at which it is released. What are the distances, in meters, traveled by the center of the ball during each second for the first 4.0 seconds of the motion?

Options:
A) 5, 10, 15, 20
B) 5, 15, 25, 35
C) 5, 20, 45, 80
D) 5, 25, 70, 150

Answer 1

The distances traveled by the center of the ball during each second for the first 4.0 seconds can be calculated using the formula d = (1/2) * g * t^2, where d is the distance traveled, g is the acceleration due to gravity, and t is the time in seconds. Using this formula, we can calculate the distances traveled as 4.9 meters, 19.6 meters, 44.1 meters, and 78.4 meters at 1, 2, 3, and 4 seconds respectively.

Step-by-step explanation:

The distance traveled by the center of the ball during each second for the first 4.0 seconds of the motion can be calculated using the formula:

1. *Formula:* d = (1/2) * g * t^2

Where:

• d is the distance traveled
• g is the acceleration due to gravity (9.8 m/s^2)
• t is the time in seconds

Using this formula, we can calculate the distances traveled during the first 4.0 seconds:

1. At t = 1 second: d = (1/2) * (9.8 m/s^2) * (1 second)^2 = 4.9 meters
2. At t = 2 seconds: d = (1/2) * (9.8 m/s^2) * (2 seconds)^2 = 19.6 meters
3. At t = 3 seconds: d = (1/2) * (9.8 m/s^2) * (3 seconds)^2 = 44.1 meters
4. At t = 4 seconds: d = (1/2) * (9.8 m/s^2) * (4 seconds)^2 = 78.4 meters