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A Comeback Post (Assignment related)
Posted: Thu Jun 24, 2021 10:54 pm
by 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 $\frac{d}{dt}(v_0 +gt) \Rightarrow g=9.8ms^{-2}$
Re: A Comeback Post (Assignment related)
Posted: Fri Jun 25, 2021 10:10 am
by 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.
Re: A Comeback Post (Assignment related)
Posted: Fri Jun 25, 2021 8:08 pm
by 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.
Re: A Comeback Post (Assignment related)
Posted: Sat Jun 26, 2021 10:43 pm
by Asif Hossain
Anindya Biswas wrote: ↑Fri Jun 25, 2021 8:08 pm
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??
Re: A Comeback Post (Assignment related)
Posted: Sat Jun 26, 2021 11:48 pm
by Anindya Biswas
Asif Hossain wrote: ↑Sat Jun 26, 2021 10:43 pm
Anindya Biswas wrote: ↑Fri Jun 25, 2021 8:08 pm
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.
Re: A Comeback Post (Assignment related)
Posted: Sun Jun 27, 2021 8:09 pm
by sakib17442
Oops, sorry I misunderstood this and voted for 0. Please forgive me
Re: A Comeback Post (Assignment related)
Posted: Fri Feb 03, 2023 3:29 pm
by celinedion
Mehrab4226 wrote: ↑Fri Jun 25, 2021 10:10 am
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!
Re: A Comeback Post (Assignment related)
Posted: Fri Sep 01, 2023 7:54 am
by otis
Mehrab4226 wrote: ↑Fri Jun 25, 2021 10:10 am
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.