Velocity equation for quadratic drag, vertically thrown

AI Thread Summary
The discussion centers on deriving the velocity equation for an object experiencing quadratic drag while moving vertically. The net force equation is established as m * dv/dt = -mg - kv^2, with initial velocity v(0) = v_0. Participants emphasize the need to apply differential equations, specifically noting that this is a first-order separable equation. Suggestions include manipulating the equation by multiplying the numerator and denominator by v and using dx = vdt. The conversation highlights the importance of understanding differential equations in solving the problem effectively.
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Homework Statement



I am having some trouble deriving this velocity equation. I the net force will be

[itеx] m * dv/dt = - mg - kv^2 [/itеx]

Because the object is moving upwards. At the time t=0 the velocity will be [itеx]v(0)=v_0[/itеx]

Homework Equations


The Attempt at a Solution

 
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Have you studied differential equations? That is a first order-separable one.

ehild
 
Try multiplying the numerator and denominator of the left hand side by v, and using dx = vdt.
 
ehild said:
Have you studied differential equations? That is a first order-separable one.

ehild

I am familiar with diff equations of first order. However i can't see what kind it is?
 
It is a first order, separable one.

ehild
 
Last edited:
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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