- #1
Ahwleung
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Homework Statement
Okay, this lab has had me stumped for the last few hours. This is our first lab for AP Physics BC.
Problem: Determine the position vector that describes the locatino of the water that is launched by a drinking fountain. Your position vector needs to be in unit vector format (i, j, k) & will include the variable t (t=0 when the water droplet is at the spout). Also determine the velocity vector, the acceleration vector, the initial velocity, the initial launch angle, and the minimum speed of the water (which is 0 at the top because the water goes up, then down)
My thoughts: It's just two dimensions so no k variable and we were only allowed to use a ruler, no stopwatches or anything else so I assume you solve for t in the equations later on. Just in case, we used cell phone timers and got t = 2.1 seconds. The Origin (0,0) is at the left edge of the drinking fountain.
The measurements are kind of complicated, so I'll just put them into a paint file.
Homework Equations
We did a problem involving vectors in which we used the equation (parenthesis = subscript)
r(f) = r(i) + v(i)*t + .5a(t^2)
So I assume you must find out r(i), v(i), change in time, and "a" (acceleration)
The Attempt at a Solution
This is the part that has me stumped. At first, the answer seems kind of obvious. Looking at the green vector, the component parts are clear (change in x = -.23 m, change in y = -.05 m) and thus the position vector is -.23i -.05j. But from there, how would one go about finding all the other answers? Heres my thoughts on the other vectors.
Velocity Vector: I'm not sure what they are asking for. Are they asking for the velocity of the water in a vector format, or the velocity of the position vector?
Acceleration Vector: Is it possible to derive a vector? Could I just derive the position vector to find velocity and derive again to find acceleration?
Initial velocity: Apparently I have to resort to the old kinematics equation to find initial velocity.
Initial Launch Angle: Now this part has me worried. I don't think you can solve for it in an equation; were we supposed to measure how HIGH the water went and do a simple tangent equation?
The minimum speed: easy, it's 0 because the velocity goes from positive to negative and thus must hit 0 at some point (like throwing a ball into the air and catching it). Or is it not that easy because the water is moving in the x direction?
Sorry for this outrageously long post. I've been staring at this lab for hours and I can't seem to make any headway into it. Thanks ahead of time for your consideration.