# When an object moves downward, is its height negative?

The question is:

A ball is thrown directly downward with an initial speed of 8.00m/s from a height of 30.0m. After what time interval does it strike the ground.

So I went through the problem and got the answer 1.79 sec. but after reviewing some notes from my teacher, I got the idea that in equations where an object is dropped down, the height has to be counted as negative. so I went through the problem again using -30m, but now my answer doesn't check out when plugged back in to x - x0 = V0 (t) + 1/2 at^2

Is there something I'm missing about this idea that height or distance is negative in drop-down questions?

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As long as your coordinate system the answer should be consistent. In this case, you did change 30m to -30m. But did you also change 8m/s to -8m/s? The same applies to the acceleration.

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so is it safe to say that for "drop down" or "free fall" questions, that the height, velocity and acceleration must all be negative? – North Sep 24 '12 at 11:34
Yes you may say that. It's really only a matter of reference point. – Xiaowen Li Sep 24 '12 at 11:46
Can you think of a reason why anyone would say that 30m off the ground is -30 by their reference point? I think my teacher was trying to describe an object moving downward, but in that case, just picturing it, the height should be positive, as well as the acceleration (because as it falls, it covers more distance each second), etc... – North Sep 24 '12 at 16:28
@North Simple. If an object is 30 m off the ground but you choose a coordinate system with its origin 60 m off the ground, then your object is 30 m BELOW the origin, which is encoded algebraically as -30 m. The negative sign really only tells you which side of the origin you're on, and YOU get to define that EVERY time you set up a problem! – user11266 Sep 24 '12 at 20:02

The way kinematics is traditionally taught leads to many misconceptions. An object's height is a property of that object and has nothing whatsoever to do with the object's motion. It's correct to speak of the object's displacement when describing motion. Displacement is a vector quantity and may have an algebraically positive or negative component depending on the coordinate system used in a problem. In introductory courses, the coordinate system chosen is usually one with its origin at Earth's surface with the positive direction directed away from Earth's surface. However, this is merely one of many choices. "Negative height" is always a meaningless term. Without specifying a coordinate system, "negative displacement" is almost as useless. The teaching community must do a better job with elementary kinematics.

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Thank you, this is what I was thinking, it makes no difference either way. Because when I do use my teacher's guidelines, the time ends up being negative, so might as well just do it from the other perspective to get a realistic answer. – North Sep 24 '12 at 16:17

In physics we typically think of an object's height as a positive number if it is higher than some reference point and negative if it's lower. For example: the height of a 3 story bulding above the ground may be about +50 feet, but the height of its basement floor would be about -15 feet. In this way yuo can calculate that the difference in height between the roof and basement floor is 50 - (-15) = 65 feet.

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If you take down as positive then displacement $s = +30$ m, initial velocity $v_i = +8$ ms$^{-1}$ and acceleration $a = +10$ ms$^{-2}$.
Using the constant acceleration kinematic equation $s = v_i t + \frac 1 2 a t^2$ where $t$ is the time gives

$$(+30) = (+8)t+\frac 1 2 (+10)t^2$$

If you take up as positive then displacement $s = -30$ m, initial velocity $v_i = -8$ ms$^{-1}$ and acceleration $a = -10$ ms$^{-2}$.
Using the constant acceleration kinematic equation $s = v_i t + \frac 1 2 a t^2$ where $t$ is the time gives

$$(-30) = (-8)t+\frac 1 2 (-10)t^2$$

the same equation as before.

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