I don't know about the DE, but

(average velocity)*time = distance

So,

(180 +400)/2 *t = 0.1

t = 0.1 /290 = 0.000345 sec -------------answer.

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Edit.

Ooops, sorry, that is wrong because the acceleration, dv/dt, is not constant.

Now I have to solve that through the DE. Heck.

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dv/dt = -kv^2 ---------------------(i)

dv /v^2 = -k dt

-1/v = -kt +C

v = 1/(kt +C) ----------------------(ii)

When v=400m/sec, t=0,

So, in (i), 400 = 1/(k*0 +C)

C = 1/400 = 0.0025

Hence, v = 1/(kt +0.0025) -----------(iii)

Develop that,

kt +0.0025 = 1/v

kt = 1/V -0.0025

kt = (1 -0.0025v)/v

k = (1 -0.0025v) /(vt) ---------------(iv)

Substitute that into (i)

dv/dt = -[(1 -0.0025v) /(vt)](v^2)

dv/dt = [(0.0025v -1)(v)] /t

dv/dt = (0.0025v^2 -v) /t

Cross multiply,

(dv)t = (0.0025v^2 -v)dt

dv /(0.0025v^2 -v) = dt /t -------------(v)

According to the Wolfram Integrator,

INT[1 /(0.0025x^2 -x)]dx = ln(400 -0.1x) -ln(x), so,

Integrate both sides of (v),

ln(400 -0.1v) -ln(v) = ln(t) +C

ln[(400 -0.1v)/v] = ln(t) +C -------------(vi)

That pesky C!

Now I'm stalled.

Heck, suppose C = 0,

[I don't know the consequence of letting C=0 here. Would the answers be wrong? I don't know. I just want to get out of this trap.]

ln[(400 -0.1v)/v] = ln(t)

(400 -0.1v)/v = t --------------------------(vii)

When the bullet comes out, the v=180 m/sec, so,

(400 -0.1*180)/180 = t

t = 382/180 = 2.12 sec.

Therefore, the bullet will pass through the wood in 2.12 seconds after it hit the wood. ----answer.