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Suppose a ball is thrown upward with a velocity of $80\frac{m}{s}$. Neglecting air resistance, How high does the ball go?

I understand that in order to solve this question, I need to use the following equation $$ v^2 = v_0^2 + 2a(y-y_0) $$ Where $v$ is final velocity, $v_0$ is initial velocity, $y$ is final position and $y_0$ is initial position.

I've looked at the solution to this question and it said the acceleration was $-9.81\frac{m}{s^2}$. I don't understand why, I read that objects in free fall, if we neglect air resistance have an acceleration of $9.81\frac{m}{s^2}$ not $-9.81\frac{m}{s^2}$.

All explanations are appreciated.

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  • $\begingroup$ Welcome to Physics! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. However, you can use the chat to discuss homework problems. $\endgroup$
    – Yashas
    Feb 24 '17 at 6:51
  • $\begingroup$ It all depends on how do you put stuff on the coordinate system. $\endgroup$
    – user79290
    Feb 25 '17 at 15:08
  • $\begingroup$ Remember that the velocities and the acceleration and positions are vectors. You must choose a coordinate system and define positive directions. If "up" is positive, the direction of the gravitational field is down, so the acceleration vector would point in the negative direction. $\endgroup$
    – Bill N
    Jan 14 '20 at 16:41
  • $\begingroup$ The choice of positive direction is arbitrary. For this problem, the author chose up as positive and down as negative. $\endgroup$ Jan 14 '20 at 17:30
  • $\begingroup$ You have 5 awesome answers! Why not accepting one of these? $\endgroup$
    – User123
    Mar 21 '20 at 21:16
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Because you're considering the upward motion. Acceleration of $-9.81 ms^{-2}$ in the upward direction means the particle is decelerating as it should be, right?

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    $\begingroup$ The direction of the motion doesnt matter. The direction of the coordinate axis does. If upward direction is positive values then the acceleration is negative and vice versa. $\endgroup$
    – lalala
    Feb 23 '17 at 21:09
  • $\begingroup$ Right. $g$ acceleration due to gravity is considered positive in the downward direction. So, when a body is thrown upward the acceleration will be $-g$. Which is the same as considering inverted coordinate axis, $y$. $\endgroup$ Feb 23 '17 at 21:15
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$v^2 - u^2 = 2*a*s$ Where $v=0$ is the final velocity $u = 80m/s$ is the initial velocity and $a=-10m/s^2$ is the acceleration of gravity. $s = -u^2/(2a)$ where $s$ is the distance traveled. So you've got acceleration in the negative direction (if you set your $y$ to go upwards), which makes a positive with this $-u^2$ and the dimensions work out. When you do any problem, always check the dimensions. Ex. if you need to find force on an object and you get it in $Newtons*meters$ or some other mumbo jumbo, you'll know you've made a mistake. P.S. If you want a want a whole derivation of the distance and velocity and acceleration, check out the Berkley's Mechanics.

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  • $\begingroup$ You have to understand that acceleration as velocity, as distance, is a vector. It has some quantity and direction. The quantity of $g=10m/s^2$ and the direction if always $down$. It all matters how you position your coordinate system. $\endgroup$ Feb 23 '17 at 22:42
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Velocity and acceleration are vectors, this means that they have both direction and magnitude.

So when I throw a ball with a certain velocity I should technically also specify the direction; after all I throw it in a certain direction, maybe along the floor or diagonally over the roof of a building; or, as in this question, straight up.

Now, when we are just interested in the magnitude of the velocity and not it's direction it's usually called speed; the same goes for acceleration except there is no usual technical term for this. Generally this is called the magnitude of the vector; and sometimes it's referred to as a scalar meaning that it's a pure number (ie not a vector).

Now, in your question/answer when it says that acceleration is 9.81*m/s*; this is a scalar value and is not specifying the direction.

When it says -9.81*m/s*, this is specifying a direction; to see this we have to specify coordinate axes; since there is no horizontal motion we can neglect that completely, so we just need to specify the vertical axis, this is just a vertical line going through where you are standing and the origin is where your feet touches the ground.

On this axis we just need to specify which is the positive direction and which is the negative direction; the simplest and the standard for throwing questions like this is to specify the upward vertical direction as positive, and hence the downward direction is negative.

Now the force of gravity on the ball acts downward; this gives an acceleration to the ball and this acceleration is also downward; this is why you have a minus in front of the acceleration, ie -9.81*m/s*.

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Both acceleration and velocity are quantities that possess direction and magnitude, for example if I kick a ball in North side with a force of 10 Newtons, I will say that I kicked that ball with a 10N force in "North" direction, see direction is important. Now when you a throw a ball in air gravity is pulling it downwards so all the effort that you put to throw the ball is being canceled out by gravity moment by moment. Due to this, you use that -ve sign, and math is telling you to use it, because if you don't how can your ball even stop, how can left hand side of your equation be 0 when you are adding non zero positive quatities on the right hand side. Hope that helps!

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This answer is dependent on how the positive y direction is defined. If you are using -9.81 as the acceleration (gravity), it is acting in the negative y-direction, which is it is negative. If you define the positive y-direction as down, you can call gravity +9.81 m/s^2. So, you can use either and will get the same answer as long as you define a positive y-direction and use that for velocity, acceleration, and change in position (y).

Hope this helps, -W

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