A body of mass m is thrown vertically into the air with an initial velocity $v_{0}$ . If the body encounters an air resistance proportional to its velocity, find (a) the equation of motion in the coordinate system of Fig. 7-6, (b) an expression for the velocity of the body at any time t, and (c) the time at which the body reaches its maximum height.

Question

Accepted Answer

(a) In this coordinate system, Eq. (7.4) may not be the equation of motion. To derive the appropriate equation, we note that there are two forces on the body: (1) the force due to the gravity given by mg and (2) the force due to air resistance given by kv, which will impede the velocity of the body. Since both of these forces act in the downward or negative direction, the net force on the body is -mg -kv. Using (7.3) and rearranging, we obtain

[latex]\frac{d v}{d t}+\frac{k}{m} v=-g[/latex] (1)

as the equation of motion.

(b) Equation (1) is a linear differential equation, and its solution is [latex]v=c e^{-(k / m) t}-m g / k ext {. At } t=0, v=v_{0}[/latex] ; hence [latex]v_{0}=c e^{-(k / m) 0}-(m g / k), ext { or } c=v_{0}+(m g / k)[/latex] . The velocity of the body at any time t is

[latex]v=\left(v_{0}+\frac{m g}{k} ight) e^{-(k / m) t}-\frac{m g}{k}[/latex] (2)

(c) The body reaches its maximum height when v = 0. Thus, we require t when v = 0. Substituting v = 0 into (2) and solving for t, we find

[latex]\begin{aligned}0 &=\left(v_{0}+\frac{m g}{k} ight) e^{-(k / m) t}-\frac{m g}{k} \e^{-(k / m) t} &=\frac{1}{1+\frac{v_{0} k}{m g}} \-(k / m) t &=\ln \left(\frac{1}{1+\frac{v_{0} k}{m g}} ight) \t &=\frac{m}{k} \ln \left(1+\frac{v_{0} k}{m g} ight)\end{aligned}[/latex]

_____________________________________________________________________________________________________________________________

Eq. (7.3) : [latex]F=m \frac{d v}{d t}[/latex]

Eq. (7.4) : [latex]\frac{d v}{d t}+\frac{k}{m} v=g[/latex]

Question 7.14: A body of mass m is thrown vertically into the air with an i...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions