# Compute the velocity of a free-falling parachutist using Euler's method for the case where m = 80 kg and c = 10 kg/s. Perform the calculation from t = 0 to 20 s with a step size of 1 s. Use an initial condition that the parachutist has an upward velocity of 20 m/s at t = 0. At t = 10 s, assume that the chute is instantaneously deployed so that the drag coefficient jumps to 50 kg/s The equation of motion is

`m*a = G - c*v`

`m*(dv)/dt = m*g -c*v`

where all quantities are positive downwards.

Taking `dt = Delta(t)= 1 s` and `dv = Delta(v) = v_(n+1) -v_n` we have

`v_(n+1) = v_n +[g -(c/m)*v_n]*Delta(t)`

Applying Euler method with the initial conditions `v_0 =-20 m/s`...

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The equation of motion is

`m*a = G - c*v`

`m*(dv)/dt = m*g -c*v`

where all quantities are positive downwards.

Taking `dt = Delta(t)= 1 s` and `dv = Delta(v) = v_(n+1) -v_n` we have

`v_(n+1) = v_n +[g -(c/m)*v_n]*Delta(t)`

Applying Euler method with the initial conditions `v_0 =-20 m/s` , `c =10 (kg)/s` we get

`v_1 =-20 +(9.8 + 10/80*20)*1 =-7.7 m/s`

`v_2 =-7.7 +(9.8+10/80*7.7)*1 = 3.06 m/s`

`v_3 =3.06 +(9.8 - 10/80*3.06) =12.48 m/s`

`v_4 =12.48+(9.8-10/80*12.48) =20.72 m/s`

`v_5 =20.72 +(9.8-10/80*20.72) =27.93 m/s`

`v_6 =27.93 +(9.8-10/80*27.93) =34.24 m/s`

`v_7 =34.24 +(9.8-10/80*34.24) =39.76 m/s`

`v_8 =39.76 +(9.8-10/80*39.76) =44.59 m/s`

`v_9 =44.59 +(9.8-10/80*44.59) =48.82 m/s`

`v_10 =48.82 +(9.8-10/80*48.82) =52.52 m/s`

At `t=10 s` the drag coefficient becomes `c =50 (kg)/s`

`v_11 =52.52+(9.8-50/80*52.52) =29.50 m/s`

`v_12 =29.50 +(9.8-50/80*29.50) =20.86 m/s`

`v_13 =20.86 +(9.8-50/80*20.86) = 17.62 m/s`

`v_14 =17.62 +(9.8-50/80*17.62) =16.41 m/s`

`v_15 =16.41 +(9.8-50/80*16.41) =15.95 m/s`

`v_16 =15.95 +(9.8 -50/80*15.95) =15.78 m/s`

`v_17 =15.78 +(9.8 -50/80*15.78) =15.72 m/s`

`v_18 =15.72 +(9.8 -50/80*15.72) =15.70 m/s`

`v_19=15.70+(9.8-50/80*15.70) =15.69 m/s`

`v_20 =15.69+(9.8-50/80*15.69) =15.68 m/s`