# one-dimension motion

A rock is thrown vertically up at 80 ft/s. Find it's maximum height, time of flight, and final velocity as it passes the starting point.

So for my parameters I have: initial y=0 final y=? acceleration=-9.8m/s^2 initial velocity=0 final velocity= 80 ft/sec time=?

So in order to find time I used the equation: v=vo +at I got t to be -8.163 seconds. I just made it positive.. so I got 8.2 seconds

To find the highest point I solved for y at the halfway time I got y to be -82.369. Should I also make this solution positive?

Are these answers correct so far? Also don't I already have the final velocity?

You got negative time because you didn't consider the direction of velocity and acceleration properly. If the increasing $y$ direction is considered positive then the initial velocity $v_0$ is positive but the acceleration is negative since it points in the minus $y$ direction.

Therefore, the equation of motion to start with is: $$y = y_0 +v_0 t + \frac{1}{2}a t^2$$

The velocity is $\frac{dy}{dt}$ or: $$\frac{dy}{dt} = v(t) = v_0 - g t$$

Where I made the substitution $a = -g$. The maximum height is marked by the velocity being zero so when $v_0 = gt$ and solving for $t$: $$t = \frac{v_0}{g}$$

The total time is twice this assuming that the rock begins to fall after $t$ seconds and the final velocity (all this ignoring air resistance) is also the same as the initial velocity of $-80 \frac{feet}{second}$ and note the minus sign for downward direction.

I compute $t$ to be $2.5$ seconds. And, after reaching the maximum height at zero velocity, the act of acceleration of gravity causes the downward velocity to increase to the value of $-g t$ or $-80 \frac{feet}{second}$.

I used $g = -32 \frac{feet}{second^2}$.

I think you made an error (possibly) in using $\frac{meters}{second^2}$ for acceleration due to gravity and did not convert your velocity from feet per second (or, maybe the mistake is that the question should have stated velocity in meters per second).

• Sorry for the subsequent edits as I left out some details. Sep 23, 2014 at 1:57

A rock is thrown vertically up at 80 ft/s. Find it's maximum height, time of flight, and final velocity as it passes the starting point.

First of all, from the instructions you provided, this here is nothing different from a projectile motion. If we consider the motion being held in a constant gravitational field, the rock acceleration would be of course the gravitational constant in said field. For instance, on earth, it will be $g=9.81\ m.s^{-1}$.

Secondly, if the rock is thrown at $80\ ft.s{-1}$, then its initial velocity is of course

\begin{equation} v_0=80\ ft.s{-1} = 24.384\ m.s^{-1} . \end{equation}

I will be working with the metric system throughout this explanation. Feel free to convert afterwards.

Then, if a starting point is not provided, it means that one can choose whatever starting point desired. For simplification purposes, let us assume that the starting position is

\begin{equation} y(t=0)=y_0=0. \end{equation}

From this information, one can simply state the equations of the rock motion by integration as follows

\begin{equation} \ddot y(t)=-g \end{equation}

\begin{equation} \dot y(t)=-gt + v_0 \end{equation}

\begin{equation} y(t)=-\frac{1}{2}gt^2 + v_0 t + y_0 =-\frac{1}{2}gt^2 + v_0 t. \end{equation}

Notice the minus sign in front of the gravitational constant/acceleration, which is there because said acceleration goes downwards (or opposed to the forward motion of the rock if you prefer).

The position $y(t)$ is a second order equation (or parabola) and we know that the highest point of said parabola is when its time derivative becomes zero. This will inform us on when does the rock attains its maximum height :

\begin{equation} \dot y(t)=0 \Leftrightarrow -gt + v_0 = 0 \Leftrightarrow t_{highest}=\frac{v_0}{g} \end{equation}

Now that we know the time when the rock is at its highest, we inject this time into the position equation $y(t)$, providing us with the maximum height :

\begin{equation} y_{max}=y(t_{highest})=-\frac{1}{2}g\left(\frac{v_0}{g}\right)^2 + v_0 \frac{v_0}{g} = \frac{{v_0}^2}{g} = 60.6095\ m. \end{equation}

The time of flight can be found simply by solving $y(t)=0$,

\begin{equation} -\frac{1}{2}gt^2 + v_0 t=0 \Leftrightarrow (-\frac{1}{2}gt + v_0)t = 0. \end{equation}

in the above equation, we have two possibilities: either $t=0$ which is obviously our starting time; or

\begin{equation} -\frac{1}{2}gt + v_0 = 0 \Leftrightarrow t_{flight}=\frac{2v_0}{g} = 4.97125\ s. \end{equation}

So now that we know that the flight duration is 4.97125 seconds or $\frac{2v_0}{g}$. We can simply use this flight time and inject it into the velocity equation, providing the velocity as the rock passes its starting position :

\begin{equation} \dot y(t_{flight})=-g \frac{2v_0}{g} + v_0 = -v_0 = -24.384\ m.s^{-1}. \end{equation}