Math Problem Statement

an object is dropped. what’s its acceleration after 1,2,3 secs

Solution

When an object is dropped in free fall near the Earth's surface, it accelerates due to gravity. The acceleration due to gravity is constant and denoted by gg, with a value of approximately 9.8m/s29.8 \, \text{m/s}^2.

The acceleration remains constant during the entire free-fall motion. Therefore, after 1, 2, 3, or any number of seconds, the object's acceleration is:

a=g=9.8m/s2a = g = 9.8 \, \text{m/s}^2

Thus, the acceleration at:

  • 1 second = 9.8m/s29.8 \, \text{m/s}^2
  • 2 seconds = 9.8m/s29.8 \, \text{m/s}^2
  • 3 seconds = 9.8m/s29.8 \, \text{m/s}^2

Since gravity is constant in free fall (ignoring air resistance), this value remains unchanged.

Would you like more details on the velocity or displacement of the object at those times?

Here are 5 related questions to consider:

  1. What is the velocity of the object after 1, 2, and 3 seconds?
  2. How far does the object fall in 1, 2, and 3 seconds?
  3. What happens to the acceleration if air resistance is considered?
  4. How does mass affect free fall acceleration?
  5. How would the acceleration change if the object were on the Moon?

Tip: The velocity of an object in free fall increases linearly with time while the acceleration remains constant.

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Math Problem Analysis

Mathematical Concepts

Physics
Kinematics
Gravitational Acceleration

Formulas

Acceleration due to gravity: a = g = 9.8 m/s²

Theorems

Free Fall Theorem
Constant Acceleration

Suitable Grade Level

Grades 9-12