Answer:

(a)

\(v=20\sqrt{2e^{-x/20}-1}.\)

(b)

\(v=20\tan\left(\frac{\pi}{4}-\frac{t}{2}\right).\)

While the particle is moving upwards, take upwards as positive. Both gravity and the resistive force act downwards. Using \(g=10\),

\(m\frac{dv}{dt}=-10m-0.025mv^2.\)

(a) Since

\(\frac{dv}{dt}=v\frac{dv}{dx},\)

we have

\(mv\frac{dv}{dx}=-10m-0.025mv^2.\)

Cancel \(m\):

\(v\frac{dv}{dx}=-(10+0.025v^2).\)

Separate variables:

\(\frac{v}{10+0.025v^2}\,dv=-dx.\)

Since \(0.025=\frac1{40}\),

\(10+0.025v^2=10+\frac{v^2}{40}=\frac{v^2+400}{40}.\)

Thus

\(\int \frac{v}{10+0.025v^2}\,dv =\int \frac{40v}{v^2+400}\,dv =20\ln(v^2+400).\)

So

\(20\ln(v^2+400)=-x+C.\)

Initially \(x=0\) and \(v=20\), so

\(C=20\ln(20^2+400)=20\ln800.\)

Hence

\(20\ln(v^2+400)=-x+20\ln800.\)

Divide by \(20\):

\(\ln(v^2+400)=\ln800-\frac{x}{20}.\)

Therefore

\(v^2+400=800e^{-x/20}.\)

So

\(v^2=800e^{-x/20}-400=400(2e^{-x/20}-1).\)

Since the particle is moving upwards, \(v\gt 0\), giving

\(v=20\sqrt{2e^{-x/20}-1}.\)

(b) From the equation of motion,

\(\frac{dv}{dt}=-(10+0.025v^2).\)

Separate variables:

\(\frac{dv}{10+0.025v^2}=-dt.\)

Using \(10+0.025v^2=\frac{v^2+400}{40}\),

\(\int\frac{dv}{10+0.025v^2} =\int\frac{40}{v^2+400}\,dv =2\tan^{-1}\left(\frac{v}{20}\right).\)

Hence

\(2\tan^{-1}\left(\frac{v}{20}\right)=-t+C.\)

Initially \(t=0\) and \(v=20\), so

\(C=2\tan^{-1}(1)=\frac{\pi}{2}.\)

Thus

\(2\tan^{-1}\left(\frac{v}{20}\right)=-t+\frac{\pi}{2}.\)

So

\(\tan^{-1}\left(\frac{v}{20}\right)=\frac{\pi}{4}-\frac{t}{2}.\)

Therefore

\(v=20\tan\left(\frac{\pi}{4}-\frac{t}{2}\right).\)