How Far Do Ink Droplets Fall in a Printer?

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Ink droplets in an ink-jet printer are ejected horizontally at a speed of 12 m/s and travel a horizontal distance of 1 mm. To determine how far they fall during this interval, the time of flight is calculated using the formula t = x/v0, resulting in approximately 8.3 x 10^-5 seconds. The vertical displacement is then calculated using the kinematic equation, considering the initial vertical velocity is zero since the ink is ejected horizontally. The resulting value for the vertical drop is negative due to the downward direction, but the absolute value is taken for the distance. The discussion highlights the importance of understanding the frame of reference in physics problems.
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Homework Statement



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?

Homework Equations


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The Attempt at a Solution



so v0 = 12 m/s

x = 1 mm.

??I am not geting the question
 
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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?
 
can you explain what they mean by " How far do they fall in this interval? "
 
i think basically its just asking for the height the ink dropped from the pen onto the paper
 
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
 
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
 
Are you sure because vy0 ( read as initial velocity of y) ? which i think should be 12
 
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 that's why the vy0 = 0 m/s
 
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
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
so should the answer be the absolute value of the answer.
 
  • #12
yes I believe so
 
  • #13
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.
 
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