Question

A shell of mass 2 kg is shot upward with an initial velocity of 200 m/sec. The magnitude of the force on the shell due to air resistance is |v|/20. When will the shell reach its maximum height above the ground? What is the maximum height

Short answer

• The shell will reach at the maximum height in 16.48 seconds
• The maximum height is 1534.8 m

Step-by-step solution

Find the velocity equation

Applying the formula for velocity$\mathrm{v}\left(\mathrm{t}\right)=\frac{\mathrm{mg}}{\mathrm{b}}+({\mathrm{v}}_{\mathrm{o}}-\frac{\mathrm{mg}}{\mathrm{b}}){\mathrm{e}}^{\frac{-\mathrm{bt}}{\mathrm{m}}}$ .

$\mathrm{v}\left(\mathrm{t}\right)=3.92.4-592.4{\mathrm{e}}^{-0.025\mathrm{t}}$ (where v_o= -200 ,m = 2, b = 1/20 ,g = 9.81)

The shell reached at the maximum height when v(t)=0 then

role="math" localid="1663929004353" $$\begin{gathered} 0=3.92.4-592.4{\mathrm{e}}^{-0.025\mathrm{t}} \\ {\mathrm{e}}^{-0.025\mathrm{t}}=\frac{392.4}{592.4} \\ {\mathrm{e}}^{-0.025\mathrm{t}}=0.66239 \\ \mathrm{t}=16.48\sec \end{gathered}$$

Hence, the shell will reach at the maximum height in 16.48 seconds

Find the equation of motion

Apply formula for value equation of motion $\mathrm{x}\left(\mathrm{t}\right)=\frac{\mathrm{mgt}}{\mathrm{b}}+\frac{\mathrm{m}}{\mathrm{b}}({\mathrm{v}}_{\mathrm{o}}-\frac{\mathrm{mg}}{\mathrm{b}}\left)\right(1-{\mathrm{e}}^{\frac{-\mathrm{bt}}{\mathrm{m}}})$

$\mathrm{x}\left(\mathrm{t}\right)=392.4\mathrm{t}-23696(1-{\mathrm{e}}^{-0.025\mathrm{t}})$

The maximum height at t = 16.48 sec then

x(t)=-1534.81m

Hence, the maximum height is 1534.8 m