# Vertical straight line motion

1. Feb 4, 2014

### jdawg

1. The problem statement, all variables and given/known data

An object is thrown vertically upward at 35 m/s. The velocity of the object 5.0 seconds later is:

2. Relevant equations

vf=vo-gt

3. The attempt at a solution
g=9.8 m/s
vo=35 m/s
t=5s
vf=?
vf=35-(9.8)(5)=14 m/s down

I know this is the correct answer, but I'm confused about why you choose the equation above. The first time I worked this problem I tried to use vf=vo+gt. Could someone please explain to me why this would be wrong? Is it maybe because the acceleration is negative?

2. Feb 4, 2014

### BvU

You're thinking the right thing. Point is: most of these equations are vector equations and vectors have a direction and a magnitude. By the time you need a number to compare with the answers in the back, you are dealing with magnitudes only.

Signs are established when you choose a coordinate system: if you say $\hat y$ (unit vector in positve y-direction) is upwards, then immediately the gravitation of the earth $\vec g$ has a negative y-component -9.81 m/s2, but he magnitude of g is still positive: 9.81 m/s2 .

The velocity vector $\vec v$ in this coordinate system has a positive y-component ("object is thrown vertically upward ") and the acceleration "increases" the y-component with $\vec g_y$ m/s every second. Here $\vec g_y = -9.81$ m/s. For short (and in the beginning that can be genuinely confusing -- you are definitely not alone there --) we disregard the vector character and write scalar equations where we take the directions into account by using the proper signs. In this case you use vf=vo-gt which is fine if you understand it. It does mean that you end up with vf < 0 . The silent understanding is that that means that vf is the y-component of the upward velocity. After all you fill in a positive v0, which still could be mistaken for a magnitude. But -14 is definitely not the magnitude of a vector.

35-(9.8)(5)=14 is not right if I ask my calculator. You correctly write "14 down", so you already do some interpreting there. Fine if you understand what you're doing, but dangerous in the beginning.

Tip: keep the units in there as long as you can:

35 m/s - 9.8 m/s2 * 5 s = - 14 m/s can be checked wrt the numbers and wrt the dimensions. Your chances to discover and fix mistakes double !

3. Feb 4, 2014

### jdawg

So would you just plug in the magnitudes into the equations? And the direction determines the sign used in the equation?

4. Feb 4, 2014

### nil1996

Just use the concept that the up is positive and down is negative.Gravity acts down so it should have a negative sign.Similarly your ball is going upwards so the velocity of ball will have a positive sign.

5. Feb 4, 2014

### jdawg

Ohhh ok, thanks guys!

6. Feb 5, 2014

### BvU

:shy:$\Biggr\downarrow$