Suppose that you toss a rock upward so that it rises and then falls back to the earth. If the acceleration due to gravity is 9.8 m/sec2, what is the rock’s acceleration at the instant that it reaches the top of its trajectory (where its velocity is momentarily zero)? Assume that air resistance is negligible.
- The acceleration of the rock is zero. - The rock has an upward acceleration of 19.6 m/s2. -The rock has a downward acceleration of 19.6 m/s2. -The rock has a downward acceleration of 9.8 m/s2. -The rock has an upward acceleration of 9.8 m/s2.
What is your initial thought, @pretty27 ?
downward acceleration of 9.8 m/s2
It will be constant during the path even if it stops for a second. But If we consider the air effect during the falling, then it CAN be zero in this case. thus it will depend on the problem you are given.
its mult choice and those are the choices given.. but i dont know.. I am second guessing this question
I think you are right, @pretty27 !
That acceleration is constant, neglecting air friction, changing gravity, and other negligible forces.
In your course, you ignore those, and acceleration is only that which is due to gravity!
ohh its downward of 9.8 m/s2 ..people is class are telling me all different answers but thanks for clearing it out1
out! *
And I mean, due to gravitational force on the mass due at around Earth's surface.
So now that this question has been answered 3 times in 24 hours, has the answer changed? :)
Haha, if you post their arguments, we can refute them if it will help you!
If the positive direction of y-axis is chosen to be upwards, then, for the time it is to fall, we have a downwards acceleration -9.8 m/s^2.
nope thanks Shane_B , theEric & Saeeddiscover for your help! just wanted to clear this out
I have one last question if you guys can help me out .. please
@Saeeddiscover , In that "up is positive" and "down is negative" system, we have downward acceleration that is \(-9.8\ [m/s^2]\). But we have a downwards acceleration with a magnitude of \(9.8\ [m/s^2]\). The choices could be more clear, but there is that one acceptable answer :)
So like you said, acceleration is \(-9.8\ [m/s^2]\) with up being positive.
:)
so - is downward and + is upward right?
The positive and negative directions can be defined however you want! But, most often, everybody says what you just said, that positive is up and negative is down.
:)
Also, feel free to post your last question when you are ready!
oki thank you
I have only 15 minutes to stay. But others are still on the opensstudy.
Yep, that is the benefit of a large community!
When an object is thrown upward, it will keep going until comes a stop. during the time fly the speed is slowing down and it would make a negative acceleration -9.8 if one of coordinates x, y, or z is directed upward. And a positive one when falling downward....
The object will slow down, yes. Now, let's suppose one axis is directed upward. Then the upward direction is positive. The change in velocity, then, would be from upwards (positive) to \(0\). The change in velocity was a negative change, and so acceleration would be negative. As the object comes down, its velocity changes from \(0\) to downward (negative). This change is also a negative change. So acceleration is still negative.
Yes it is a correct context. I think it will stay negative for both directions. Can you see a symmetric in direction?
A symmetry in direction?
The acceleration is negative for both directions whereas the velocity vector changes two times during the flying. it is a dramatic motion!
Ah, I was wondering if that was what you meant! Yes, the change in position is first positive and then negative! All while position is positive and acceleration is negative! It's fun to think about!
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