Vertical motion of an object We define acceleration as -9.8(m)/(s^2) for an object that is thrown upwards due to gravity. My question is when the object reaches maximum height at the point where it is about to turn around downwards, will the magnitude of the acceleration still be 9.8m/(s^2) or will it be greater than this as to help it turn around?

Kayden Mills

Kayden Mills

Answered question

2022-11-07

Vertical motion of an object
We define acceleration as 9.8 m s 2 for an object that is thrown upwards due to gravity. My question is when the object reaches maximum height at the point where it is about to turn around downwards, will the magnitude of the acceleration still be 9.8 m s 2 or will it be greater than this as to help it turn around?

Answer & Explanation

klofnu7c2

klofnu7c2

Beginner2022-11-08Added 14 answers

It will be equal to 9.8 m s 2 . Acceleration due to gravity is constant, as long as it doesn't stray far away from the surface of the Earth.
kaltEvallwsr

kaltEvallwsr

Beginner2022-11-09Added 8 answers

I'm guessing your confusion is based on a misconception about acceleration and velocity. Acceleration is the rate of change of velocity with time. So as time passes, velocity should change at a certain rate. This rate is the acceleration. So as the ball decelerates, since gravity is acting against it, it's velocity should vary. Hence the velocity of the ball at Max height would be zero, but it's acceleration is still constant. Acceleration can be constant (in the case of gravity) and variable. Variable acceleration simply means that the rate at which velocity changes is not constant i.e it varies. For example a bus driver's acceleration is variable, because he would always have to decelerate at some point.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Force, Motion and Energy

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?