Finding the magnitude of velocity given vx and vy

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To find the magnitude of velocity given the components vx and vy, use vector addition based on Pythagorean theorem. The equation v² = vx² + vy² allows for calculating the hypotenuse of a right triangle formed by the velocity components. The discussion clarifies that this problem does not require calculus, as it involves basic vector addition. The participant initially misunderstood the question but confirmed their understanding after receiving guidance. This approach effectively resolves the query regarding calculating velocity magnitude.
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Homework Statement


I am working on a lab, and I have come up with graphs (and the data set) for vy vs. time and vx vs. time, but I'm not sure how to determine v? I just need to come up with the equation so solve for v, but we haven't done anything calc oriented in the course yet.


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The Attempt at a Solution

 
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It has nothing to do with calculus, just simple vector addition. The horizontal component of v is vx while the vertical component of v is vy. Hence, by Pythagora's theorem, we have v^{2} = v_{x}^{2} + v_{y}^{2}
 
Have you seen vectors? There is a right triangle with vx as one right side and vy as the other. You are looking for the speed v which is the hypotenuse of the triangle. So ...
 
Okay thank you both for your help.
I guess I just misunderstood the question, we have covered adding vectors in the course so I am able to do that.

thanks again!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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