Recent content by Some1WhoNeedsHelp

I am stuck at A, how do you calculate the motion of the air? I know ρ=m/V but what is the mass of the air, or I got that question wrong? Thanks

you help me a lot to understand too, even with this guy help I wouldn't be able to understand what he tries to explain to me if not for you two That is why I am thanking you Let me just upload my homework and I will solve this

I found someone in real life who help me to better understand the subject. I realize now that the arrow won't stop as long as only drag force is active and gamma is 1/meter Still thanks to both of you.

Alright so according to the equation I created from post #21 -1/(γvt)+1/(γv0)=t -v0+vt=-γvtv0t vt+γvtv0t=v0 vt(1+γv0t)=v0 vt=v0/(1+γv0t) So as time keeps increasing the vt will decrease to the point the speed of the arrow will become neglectable, so the arrow will stop

Oh wait, Is that mean I need to find an expression of V depending on t? So am I right about the drag force will stop the arrow eventually despite being the only force?

-1/(γvt)+1/(γv0)=t I know γ,V0 so I get - 1 equation with 2 unknown things (Vt,t) I would assume Vt should be equal to 0 since the drag force will stop the arrow eventually but the question asks me to calculate the value of Vt using the equation.I also know η, Cd , A, ρ and the length of the...

The integral of (1/x2).dx is (-1/x) so (-1/γv0)-(-1/γv)=-t

∫v0v(1/γV2) dv=∫0tdt (v0/γ)*(-1/v0)-(v/γ)*(-1/v)=-t (-1/γ)-(-1/γ)=t 0=t correct?

yes! But since γ is positive I have a new problem I am trying to calculate the V so I multiply by "dt" and do integral ∫(1/γV2) dv=∫dt (v/γ)*(-1/v)=t (-1/γ)=t So time is negative

so γ=(Cd*ρ*A)/2m ?

Because the force is opposite to direction of motion, the equation needs to be negative? F=-(Cd*ρ*A*V2)/2=-mγV2? so γ=(Cd*ρ*A)/2m

So should I do Absolute value?F=ma F=(Cd*ρ*A*V2)/2 ma=(Cd*ρ*A*V2)/2 a=(Cd*ρ*A*V2)/2m a = dv/dt dv/dt = -γV2 -γV2=(Cd*ρ*A*V2)/2m γ=|-(Cd*ρ*A)/2m|

a = dv/dt acceleration is meter/(seconds)2 and v2 is (meter)2/(seconds)2 So gamma must be meter and using Newton's law ma=(Cd*ρ*A*V2)/2 and since a = dv/dt Gamma=-(Cd*ρ*A)/2mIs that correct?

it's a function of speed relative to the time it represents The change in speed is divided by the change in time. So does it represent the location of the arrow? Is that got something to do with the function: X=X0+VT+(1/2)AT2 ?

1=-gamma v^2? I also need to find gamma, So I need to realize what the letter supposed to represent I looked at the formula sheet and didn't find it there