1. Feb 3, 2009

### tnutty

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

Ink droplets in an ink-jet printer are ejected horizontally at 12 and travel a horizontal distance of 1.0 to the paper.

How far do they fall in this interval?

2. Relevant equations
?

3. The attempt at a solution

so v0 = 12 m/s

x = 1 mm.

??I am not geting the question

2. Feb 3, 2009

### needlottahelp

umm...i think you use
x = (v0) (t) and solve for time

then use the time for one of the kinematics equation as if it were free fall? does that make sense to you?

3. Feb 3, 2009

### tnutty

can you explain what they mean by " How far do they fall in this interval? "

4. Feb 3, 2009

### needlottahelp

i think basically its just asking for the height the ink dropped from the pen onto the paper

5. Feb 3, 2009

### tnutty

so would this be correct :

x = v0 *t.

t = x/v0

= 1mm / 12 m/s

= 0.001m / 12 m/s

t = 8.3 x 10^-5

so using the equation,

Y = Y0 +Vy0*t -1/2 g*t^2

y = 0 + 12(8.3 x 10^-5) - 1/2 (9.8) (8.3 x 10 ^-5)^2

= 10.0 x 10^-4

6. Feb 3, 2009

### needlottahelp

I see no problem with the time. The equation you used was correct too but the problem is that vyo = 0 m/s and not twelve because the angle of elevation from the horizontal is 0 therefore vyo = sin (0 degrees) (12 m/s) = 0 m/s

7. Feb 3, 2009

### tnutty

Are you sure because vy0 ( read as initial velocity of y) ? which i think should be 12

8. Feb 3, 2009

### needlottahelp

well vy0 is the initial velocity in the y-direction. In other words the vertical direction. The ink was ejected horizontally 12 m/s so thats why the vy0 = 0 m/s

9. Feb 3, 2009

### tnutty

x = v0 *t.

t = x/v0

= 1mm / 12 m/s

= 0.001m / 12 m/s

t = 8.3 x 10^-5

so using the equation,

Y = Y0 +Vy0*t -1/2 g*t^2

y = 0 + 0(8.3 x 10^-5) - 1/2 (9.8) (8.3 x 10 ^-5)^2

= -3.40 x 10^-8....

The number should not be negative?

10. Feb 3, 2009

### needlottahelp

Its negative because of the frame of reference. remember displacement is a vector? its only negative because its downward. the opposite would be true
the distance is just 3.40 x 10^-8

11. Feb 3, 2009

### tnutty

12. Feb 3, 2009

### needlottahelp

yes I believe so

13. Feb 3, 2009

### tnutty

Thank you for you help. As you can see I am very new to this and also having a hard time grasping the method and concept of solving physics problem. Although I do not know why, because I am relatively good at quantitative work.