Answer:
[tex]v_{max}=52.38\frac{m}{s}[/tex]
[tex]v_{100}=33.81[/tex]
Explanation:
the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:
[tex]\sum{F}=0=F_d-W[/tex]
[tex]F_d=W[/tex]
[tex]kv_{max}^2=m*g[/tex]
[tex]v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}[/tex]
To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:
[tex]\sum{F}=ma=W-F_d[/tex]
[tex]ma=W-F_d[/tex]
[tex]ma=mg-kv_{100}^2[/tex]
[tex]a=g-\frac{kv_{100}^2}{m}[/tex] (1)
consider the next equation of motion:
[tex]a = \frac{(v_{x}-v_0)^2}{2x}[/tex]
If assuming initial velocity=0:
[tex]a = \frac{v_{100}^2}{2x}[/tex] (2)
joining (1) and (2):
[tex]\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}[/tex]
[tex]\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g[/tex]
[tex]v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g[/tex]
[tex]v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}[/tex]
[tex]v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}[/tex] (3)
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}[/tex]
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}[/tex]
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}[/tex]
[tex]v_{100}=\sqrt{1,143.3}[/tex]
[tex]v_{100}=33.81[/tex]
To plot velocity as a function of distance, just plot equation (3).
To plot velocity as a function of time, you have to consider the next equation of motion:
[tex]v = v_0 +at[/tex]
as stated before, the initial velocity is 0:
[tex]v =at[/tex] (4)
joining (1) and (4) and reducing you will get:
[tex]\frac{kt}{m}v^2+v-gt=0[/tex]
solving for v:
[tex]v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }[/tex]
Plots:
