Answer:
Explanation:
Given
acceleration is given by
[tex]a=-g-c_Dv^2[/tex]
where [tex]\ddot{y}=a[/tex]
[tex]\dot{y}=v[/tex]
Also acceleration is given by
[tex]a=v\frac{\mathrm{d} v}{\mathrm{d} s}[/tex]
[tex]ds=\frac{v}{a}dv[/tex]
[tex]\int ds=\int \frac{v}{-g-0.001v^2}dv[/tex]
[tex]\Rightarrow Let -g-0.001v^2=t[/tex]
[tex]-0.001\times 2vdv=dt[/tex]
[tex]vdv=-\frac{dt}{0.002}[/tex]
[tex]at\ v_0=50\ m/s,\ t=-g-0.001(50)^2[/tex]
[tex]t=-g-2.5[/tex]
at [tex]v=0,\ t=-g[/tex]
[tex]\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}[/tex]
[tex]\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}[/tex]
[tex]s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}[/tex]
[tex]s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})[/tex]
[tex]s=113.608\ m[/tex]
when air drag is neglected maximum height reached is
[tex]h=\frac{v_0^2}{2g}[/tex]
[tex]h=\frac{50^2}{2\times 9.8}[/tex]
[tex]h=127.55\ m[/tex]
