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1. A parachutist of mass

The parachutist falls from rest freely under gravity for a time [itex]\frac{m}{2k}[/itex] and then opens his parachute. Prove that the total distance he has fallen when his velocity is [itex]\frac{3mg}{4k}[/itex] is [itex]\frac{m^2 g}{8 k^2} (8 \ln2 - 1)[/itex]

2.

[tex]a = v \frac{dv}{ds} [/tex]

[tex] v^2 = u^2 + 2 a s [/tex]

[tex] v = u + a t [/tex]

3. Until parachute is opened:

[tex]t = \frac{m}{2k}[/tex]

[tex]a = g[/tex]

[tex]u = 0[/tex]

[tex]v = u + a t[/tex]

[tex]v = \frac{mg}{2k}[/tex]

[tex] v^2 = u^2 + 2 a s [/tex]

[tex] \frac{m^2 g^2}{4 k^2} = 2 g s [/tex]

[tex] \frac{m^2 g}{8 k^2} = s [/tex]

When parachute is opened:

Force downwards [itex]ma = mg - kv[/itex]

[tex] m v \frac{dv}{ds} = mg - kv [/tex]

[tex] \int \frac{mv}{mg - kv} dv = \int ds[/tex]

Substitution: [tex] u = mg - kv [/tex]

[tex]\frac{du}{dv} = -k[/tex]

[tex]v = \frac{u - mg}{-k}[/tex]

[tex] \int \frac{mv}{-k (mg - kv)} du = \int ds[/tex]

[tex] \int \frac{m(u - mg)}{u} du = \int ds[/tex]

[tex] \int m - \frac{m^2 g}{u} du = \int ds[/tex]

[tex] mu - m^2 g \ln u = \int ds[/tex]

[tex] m (mg - kv) - m^2 g \ln (mg - kv) = s + c[/tex] where c is a constant

To work out constant:

[tex]v = \frac{mg}{2k}[/tex]

[tex]s = \frac{m^2 g}{8 k^2}[/tex]

[tex]m^2 g - mk \frac{mg}{2k} - m^2 g \ln ( mg - \frac{kmg}{2k} ) = \frac{m^2 g}{8 k^2} + c[/tex]

[tex]\frac{m^2 g}{2} - m^2 g \ln ( \frac{mg}{2} ) - \frac{m^2 g}{8 k^2} = c[/tex]

Putting c back into equation:

[tex]m^2 g - mkv - m^2 g \ln ( mg - kv ) - \frac{m^2 g}{2} + m^2 g \ln ( \frac{mg}{2} ) + \frac{m^2 g}{8 k^2} = s[/tex]

Putting in value of v:

[tex]- \frac{m^2 g}{4} + m^2 g \ln 2 + \frac{m^2 g}{8 k^2} = s[/tex]

[tex]\frac{m^2 g}{8} ( 8 \ln 2 - 2 + \frac{1}{k^2} ) = s[/tex]

So i have wrong answer where have i gone wrong?

*m*falls freely until his parachute opens. When it is open he experiences an upward resistance*kv*where*v*is his speed and*k*is a positive constant.The parachutist falls from rest freely under gravity for a time [itex]\frac{m}{2k}[/itex] and then opens his parachute. Prove that the total distance he has fallen when his velocity is [itex]\frac{3mg}{4k}[/itex] is [itex]\frac{m^2 g}{8 k^2} (8 \ln2 - 1)[/itex]

2.

[tex]a = v \frac{dv}{ds} [/tex]

[tex] v^2 = u^2 + 2 a s [/tex]

[tex] v = u + a t [/tex]

3. Until parachute is opened:

[tex]t = \frac{m}{2k}[/tex]

[tex]a = g[/tex]

[tex]u = 0[/tex]

[tex]v = u + a t[/tex]

[tex]v = \frac{mg}{2k}[/tex]

[tex] v^2 = u^2 + 2 a s [/tex]

[tex] \frac{m^2 g^2}{4 k^2} = 2 g s [/tex]

[tex] \frac{m^2 g}{8 k^2} = s [/tex]

When parachute is opened:

Force downwards [itex]ma = mg - kv[/itex]

[tex] m v \frac{dv}{ds} = mg - kv [/tex]

[tex] \int \frac{mv}{mg - kv} dv = \int ds[/tex]

Substitution: [tex] u = mg - kv [/tex]

[tex]\frac{du}{dv} = -k[/tex]

[tex]v = \frac{u - mg}{-k}[/tex]

[tex] \int \frac{mv}{-k (mg - kv)} du = \int ds[/tex]

[tex] \int \frac{m(u - mg)}{u} du = \int ds[/tex]

[tex] \int m - \frac{m^2 g}{u} du = \int ds[/tex]

[tex] mu - m^2 g \ln u = \int ds[/tex]

[tex] m (mg - kv) - m^2 g \ln (mg - kv) = s + c[/tex] where c is a constant

To work out constant:

[tex]v = \frac{mg}{2k}[/tex]

[tex]s = \frac{m^2 g}{8 k^2}[/tex]

[tex]m^2 g - mk \frac{mg}{2k} - m^2 g \ln ( mg - \frac{kmg}{2k} ) = \frac{m^2 g}{8 k^2} + c[/tex]

[tex]\frac{m^2 g}{2} - m^2 g \ln ( \frac{mg}{2} ) - \frac{m^2 g}{8 k^2} = c[/tex]

Putting c back into equation:

[tex]m^2 g - mkv - m^2 g \ln ( mg - kv ) - \frac{m^2 g}{2} + m^2 g \ln ( \frac{mg}{2} ) + \frac{m^2 g}{8 k^2} = s[/tex]

Putting in value of v:

[tex]- \frac{m^2 g}{4} + m^2 g \ln 2 + \frac{m^2 g}{8 k^2} = s[/tex]

[tex]\frac{m^2 g}{8} ( 8 \ln 2 - 2 + \frac{1}{k^2} ) = s[/tex]

So i have wrong answer where have i gone wrong?

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