How do I know when velocity is negative in a 1-D motion problem?

  • #1
1
0

Homework Statement


Two baseballs are set in motion from the top of a 5 m tall tower. One is thrown straight down toward the ground with an initial velocity of 20 m/s. The second is thrown straight up toward the sky with an initial velocity of 20 m/s.

How do their velocities compare when they hit the ground?

Homework Equations


Vx^2=Vx0^2 + 2*a(x2-x1)

The Attempt at a Solution


This what my teacher wrote:

#1 (Ball thrown straight down)
Vx0= 20 m/s
Vx=?
t=?
x=5 m
a=10 m/s^2

Vx^2=Vx0^2 + 2*a(x2-x1)
Vx^2=500
Vx = 22.36 m/s

#2 (Ball thrown straight up)
Vx0= -20 m/s
Vx= ?
a= 10 m/s^2
x= 5 m

Vx^2=Vx0^2 + 2*a(x2-x1)
Vx^2=(-20)^2 + 2(10)(5)
Vx^2=500
Vx=22.36 m/s

I don't understand why the initial velocity for the ball thrown straight up is negative. Also, since it's downward acceleration, wouldn't the acceleration for the ball thrown up be negative (it's going AGAINST gravity, correct?).
 

Answers and Replies

  • #2
416
51
Basically, being negative doesn't mean that it is going against something. Neither being positive is going with something (Not always at least)
You should just put a coordinate system and your system here is just defined as up is negative and down is positive.

So answer these questions:
1) What is the sign of the velocity of the first ball?
2) What is the sign of the velocity of the second ball?
3) acceleration is a vector too, What is the direction of it ? Knowing that will tell you the sign of it.
 
  • #3
PeroK
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Homework Statement


Two baseballs are set in motion from the top of a 5 m tall tower. One is thrown straight down toward the ground with an initial velocity of 20 m/s. The second is thrown straight up toward the sky with an initial velocity of 20 m/s.

How do their velocities compare when they hit the ground?

Homework Equations


Vx^2=Vx0^2 + 2*a(x2-x1)

The Attempt at a Solution


This what my teacher wrote:

#1 (Ball thrown straight down)
Vx0= 20 m/s
Vx=?
t=?
x=5 m
a=10 m/s^2

Vx^2=Vx0^2 + 2*a(x2-x1)
Vx^2=500
Vx = 22.36 m/s

#2 (Ball thrown straight up)
Vx0= -20 m/s
Vx= ?
a= 10 m/s^2
x= 5 m

Vx^2=Vx0^2 + 2*a(x2-x1)
Vx^2=(-20)^2 + 2(10)(5)
Vx^2=500
Vx=22.36 m/s

I don't understand why the initial velocity for the ball thrown straight up is negative. Also, since it's downward acceleration, wouldn't the acceleration for the ball thrown up be negative (it's going AGAINST gravity, correct?).
The simple answer is that you get to choose what direction is positive and which direction is negative. You either choose up to be positive or down to be positive. Your teacher chose down to be positive, but you could equally well solve the problem choosing up to be positive.

If you choose up to be positive, then gravity is a negative acceleration (downwards).

If you choose down to be positive, then gravity is a positive acceleration.
 

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