A Comeback Post (Assignment related)

[Asif Hossain]

For HSC first year in the first week (surely everybody has submitted that by now) there was a problem stating:
What is the acceleration of a vertically thrown object at the peek of its path?
Somebody say it is $0ms^{-2}$ some say it is $9.8ms^{-2}$
I am on the side of $9.8ms^{-2}$ here is my reason:
At the peek the object is going to be a free falling object so at time $t$ its velocity $v=v_0 +gt$
So, at any moment the acceleration should be $\frac{d}{dt}(v_0 +gt) \Rightarrow g=9.8ms^{-2}$

[Mehrab4226]

I am also on the side of $9.8 ms^{-2}$, because if the acceleration is $0$. Then the object is not supposed to move down. But it does move down. So it has acceleration.

[Anindya Biswas]

Acceleration is the second derivative of position with respect to time.
The position of the object is given by,
$y(t)=v_0t-\frac12gt^2$
$\Rightarrow y''(t)=-g$

Some confusion is arising that since the object was moving upwards, there should be an upward force acting on that object, and the gravitational force cancelled that force and the total force should be zero.
But the problem is, there was no force acting on that thing upwards. The upward force acted when it was in our hand. As soon as it left the ground, all upward forces stopped acting and the object moved upward for its inertia. So gravitational force never cancelled anything, it was the only force acting there.

[Asif Hossain]

Acceleration is the second derivative of position with respect to time.
The position of the object is given by,
$y(t)=v_0t-\frac12gt^2$
$\Rightarrow y''(t)=-g$

Some confusion is arising that since the object was moving upwards, there should be an upward force acting on that object, and the gravitational force cancelled that force and the total force should be zero.
But the problem is, there was no force acting on that thing upwards. The upward force acted when it was in our hand. As soon as it left the ground, all upward forces stopped acting and the object moved upward for its inertia. So gravitational force never cancelled anything, it was the only force acting there.

-g or g??

[Anindya Biswas]

Acceleration is the second derivative of position with respect to time.
The position of the object is given by,
$y(t)=v_0t-\frac12gt^2$
$\Rightarrow y''(t)=-g$

Some confusion is arising that since the object was moving upwards, there should be an upward force acting on that object, and the gravitational force cancelled that force and the total force should be zero.
But the problem is, there was no force acting on that thing upwards. The upward force acted when it was in our hand. As soon as it left the ground, all upward forces stopped acting and the object moved upward for its inertia. So gravitational force never cancelled anything, it was the only force acting there.

-g or g??

Well, the position function was upward displacement, so $-g$ is the upward acceleration. That means $g$ is the downward acceleration.

[sakib17442]

Oops, sorry I misunderstood this and voted for 0. Please forgive me

[celinedion]

I am also on the side of 98ms−2, because if the acceleration is 0 . Then the object is not supposed to move down. But it does move down. So it has acceleration.mapquest driving directions

Me too!

[otis]

I am also on the side of $9.8 ms^{-2}$, because if the acceleration is $0$. Then the object is not supposed to move down. But it does move down. So it has acceleration. slope game only up

That's right, I'm thinking like this.