Question

A ball with a weight of 20 N is thrown vertically upward . What is the acceleration of the ball just as it reaches the top of its path ?
A. $10m{s^{ - 2}}$downward
B. $10m{s^{ - 2}}$upward
C. $20m{s^{ - 2}}$downward
D. $20m{s^{ - 2}}$upward
E. Zero

Answer 1

The net acceleration given to objects by the combined action of gravitation (from mass distribution within Earth) and centrifugal force (from the Earth's rotation) is represented by the letter g. Newton's second law of motion, or F = ma (force = mass x acceleration), defines the weight of an item on Earth's surface as the downward force on that object.

Complete step by step solution:
The downward acceleration of a free-falling object is $9.8{\text{ }}m{s^{ - 2}}$(on Earth). This numerical number for a free-falling object's acceleration is so significant that it is given a specific name. It's called the acceleration of gravity, and it's the acceleration of any object travelling only due to gravity. In fact, the acceleration of gravity is such a significant number that scientists have given it its own symbol, g. throughout its entire motion, the only acceleration acting on that body is gravity's acceleration. Gravity acceleration is a vector quantity with both magnitude and direction. Gravity would point directly towards the sphere's centre in a spherically symmetric Earth. Because the Earth's shape is significantly flatter, substantial variations in gravity direction exist: basically the difference between geodetic and geocentric latitude. Local mass anomalies, such as mountains, produce smaller deviations, known as vertical deflection.
Hence option A $10m{s^{ - 2}}$downward is correct

Note: $9.8{\text{ }}m{s^{ - 2}}$is the most precise numerical figure for the acceleration of gravity. This numerical number (to the second decimal place) has minor fluctuations that are largely determined by altitude. In The Physics Classroom Tutorial, we will occasionally utilise the estimated value of ${\text{10 }}m{s^{ - 2}}$to simplify the various mathematical activities that we will complete with this quantity. We will be able to better focus on the conceptual nature of physics as a result of this, without sacrificing too much numerical precision.