# PGRE8677 Problem 1

A rock is thrown vertically upward with initial speed $$v_0$$. Assume a friction force proportional to $$-\mathbf{v}$$, where $$\mathbf{v}$$ is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the following is correct?

A. The acceleration of the rock is always equal to $$\mathbf{g}$$.

B. The acceleration of the rock is equal to $$\mathbf{g}$$ only at the top of the flight.

C. The acceleration of the rock is always less then $$\mathbf{g}$$.

D. The speed of the rock upon return to its starting point is $$v_0$$.

E. The rock can attain a terminal speed greater than $$v_0$$ before it returns to its starting point.

## Solution

With falling objects on Earth, we can model them pretty simply. If there was no wind resistance, then we can model the position as,

$y(t) = y_0 + v_0 t - \frac{1}{2}gt\text{,}$

where $$y(t)$$ is the vertical position at time, $$t$$, $$y_0$$ is the initial position, $$v_0$$ is the initial velocity (positive if we are throwing the object upward), and g is the acceleration due to gravity, $$9.80665$$.

Let $$S \subseteq \mathbb R$$ where $$y \in S$$ is the position. Since we're working in $$\mathbb R$$, $$\mathbf{y}$$ has direction of $$+$$ or $$-$$. We can provide any initial position along $$\mathbb R$$, the initial velocity scaled by $$t$$ and since gravity is acting upon the object, $$g$$ is the acceleration due to gravity which is scaled by the square of $$t$$ and pulling the object back down towards the Earth. However, this isn't what we experience in the "real world" since we are dealing with wind resistance. Let's approach this a little differently.

The total force is given by,

$\sum \mathbf{F} = \mathbf{F}_\text{g} + \mathbf{F}_\text{air} = mg - ks(t)\text{,}$

which gives us the differential equation,

$mg - k\dot{y}(t) = m\ddot{y}(t)\text{,}$

where $$m$$ is the mass of the object, $$g$$ is still the acceleration due to gravity, $$k$$ is a constant of proportionality of the object, $$y$$ is the position, $$\dot{y}$$ is the velocity, and $$\ddot{y}$$ is the acceleration.

We can re-write (3) as,

$$\frac{mg}{k} - \dot{y}$$ $\ddot{y}(t) = \frac{k}{m}\bigg((t)\bigg)$

We can introduce an auxilliary function, $$p(t) = \frac{mg}{k} - y(t)$$ which means that $$\dot{p}(t) = -\dot{y}(t)$$. Exclusively in terms of $$p(t)$$,

$-\dot{p}(t) = \frac{k}{m}p(t)$