Question

A projectile of mass \(m\) is launched vertically upward from ground level at time \(t=0\) with initial velocity \(v_{0}\) and is acted upon by gravity and air resistance. Assume the drag force is proportional to velocity, with drag coefficient \(k\). Derive an expression for the time, \(t_{m}\), when the projectile achieves its maximum height.

Short answer

Based on the given step-by-step solution, derive an expression for the time when the projectile reaches its maximum height, assuming air resistance is proportional to velocity.

Step-by-step solution

Express forces acting on the projectile

Since the only forces acting on the object are air resistance and gravitational force, we can write the net force acting on the projectile as: $$F_{net} = F_{drag} + F_{gravity}$$ Since we are told that the drag force is proportional to velocity, we can write \(F_{drag} = -kv\) where \(k\) is the drag coefficient and \(v\) is the velocity. Gravitational force acting on the projectile is \(F_{gravity} = -mg\). So, the net force becomes: $$F_{net} = -kv - mg$$

Set up the differential equation

Using Newton's second law, we can write \(F_{net} = m\times a\), where \(a\) is the acceleration. As acceleration is the time derivative of velocity, we have: $$m\frac{dv}{dt} = -kv - mg$$ We have a first-order linear differential equation.

Solve the differential equation for velocity

To solve the differential equation, we can use an integrating factor. The integrating factor is given by \(e^{\int \frac{k}{m}dt}=e^{\frac{kt}{m}}\). Multiply equation by integrating factor on both sides: $$me^{\frac{kt}{m}}\frac{dv}{dt}=-ke^{\frac{kt}{m}}v-mge^{\frac{kt}{m}}$$ Now, this becomes an exact differential equation as: $$\frac{d}{dt}(ve^{\frac{kt}{m}})=-ke^{\frac{kt}{m}}v-mge^{\frac{kt}{m}}$$ Integrate both sides with respect to \(t\): $$\int d(ve^{\frac{kt}{m}}) = \int(-ke^{\frac{kt}{m}}v - mge^{\frac{kt}{m}}) dt$$ After integrating, we get: $$ve^{\frac{kt}{m}} = \int(-ke^{\frac{kt}{m}}v - mge^{\frac{kt}{m}}) dt$$

Find velocity as a function of time

To find the expression for velocity \(v\) as a function of time \(t\), we solve for \(v\): $$v(t) = e^{-\frac{kt}{m}} \left[\int (-ke^{\frac{kt}{m}}v - mge^{\frac{kt}{m}}) dt\right]$$

Determine when the velocity becomes zero

To find the time \(t_m\) when the projectile reaches its maximum height, we set the velocity to zero and solve for \(t\): $$0 = e^{-\frac{kt}{m}} \left[\int (-ke^{\frac{kt}{m}}v - mge^{\frac{kt}{m}}) dt\right]$$ Solving for \(t\), we get the time \(t_m\) when the projectile reaches its maximum height.