# Homework Help: 1D kinematics - finding g from time traveled

1. Apr 13, 2013

### swevener

1. The problem statement, all variables and given/known data
This is from Kleppner, problem 1.12 in the 2010 edition, if you want to see the diagram.

The acceleration of gravity can be measured by projecting a body upward and measuring the time that it takes to pass two given points in both directions. Show that if the time the body takes to pass a horizontal line $A$ in both directions is $T_A$, and the time to go by a second line $B$ in both directions is $T_B$, then, assuming that the acceleration is constant, its magnitude is
$$g = \frac{8 h}{T_A^2 - T_B^2},$$where $h$ is the height of line $B$ above line $A$.

2. Relevant equations
$y = y_0 + \dot{y}_0 t - \frac12 g t^2$

$\dot{y} = \dot{y}_0 - g t$

3. The attempt at a solution
Since we're only interested in what happens above height $A$, I'm going to make that my zero height, so the total flight time is $T_A$, making the time for the body to reach its max height $T_A / 2$. From that I can get the initial velocity from $\dot{y} = 0 = \dot{y}_0 - g ( T_A / 2 )$, giving me
$$\dot{y}_0 = \frac{g T_A}{2}.$$ So now I've got an equation for height
$$y = \frac{g}{2} ( T_A t - t^2 ),$$which I can use to find an expression for the max height by plugging in $t = T_A / 2$:
$$y_{\text{max}} = \frac{g}{2} \left( T_A \frac{T_A}{2} - \frac{T_A^2}{4} \right) = \frac{g T_A^2}{8}.$$
I also want to know how long it takes to get to $B$; I'll call that $t_h$. Plugging that into my height equation,
$$h = \frac{g}{2} ( T_A t_h - t_h^2 ),$$ and solving for $t_h$ I get
$$t_h = \frac{T_A}{2} \pm \frac12 \sqrt{T_A^2 - \frac{8 h}{g}},$$ the minus term being the time it takes to reach $B$ the first time and plus term the time it takes to reach $B$ the second time.

Now I'm going to focus on $B$. I can get another expression for the max height by using time $T_B/2$ and initial velocity $\dot{y}_B = \dot{y}_0 - g t_h$. Plugging in everything I have so far,
$$\dot{y}_B = \frac{g T_A}{2} - \left( \frac{g T_A}{2} - \frac12 \sqrt{T_A^2 - \frac{8 h}{g}} \right) = \frac12 \sqrt{T_A^2 - \frac{8 h}{g}}.$$And the equation for max height is
$$y_{\text{max}} = \frac{T_B}{2} \frac12 \sqrt{T_A^2 - \frac{8 h}{g}} - \frac{g}{2} \frac{T_B^2}{4} = \frac{T_B}{4} \sqrt{T_A^2 - \frac{8 h}{g}} - \frac{g T_B^2}{8} = \frac{g T_A^2}{8}.$$Simplifying,
$$T_B \sqrt{T_A^2 - \frac{8 h}{g}} = \frac{g}{2} ( T_A^2 + T_B^2 ).$$
Here's where I'm stymied. If there's a way to solve that for $g$, I don't know it, and the magic equation-solving machines don't spit out anything nice, certainly not the given solution. So I suppose I've gone astray somewhere. Help?

2. Apr 13, 2013

### Staff: Mentor

Looks overly complicated.

Hint: View the body as it falls, starting from the highest point.

3. Apr 13, 2013

### swevener

Solved it. Thanks.