# Vertical straight line motion

## Homework Statement

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

vf=vo-gt

## 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?

BvU
Homework Helper
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 !

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

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.

Ohhh ok, thanks guys!

BvU