New Year, New Heights and the Projectile

I spent my new year celebration at our friend’s place with 2 other colleagues. It was started with a dinner at about 9:30 pm and we bought some bottles of wine in addition to the spirits we already brought. At about 10:30, a street party was setup in the vicinity and about 5 other families joined in a potluck. Marijo’s brother started the fireworks at about that time with David and Gerard (my colleagues) taking turns setting them off. I did not attempt to set them off because I did a lot those things way back as a child and still lucky enough to have my fingers intact.

This is a video of our 2012 New Year’s Celebration:

As Gerard was flying off skyrockets (or kwitis as they are called in my language) David asked me “How high do you think these things go up?”. That was a really good question which got me to thinking. I said “Let’s compute!” So I took my iphone and launched its built-in stopwatch. Gerard flew about 3 skyrockets and I took the time it takes for them to fly until they blow up in the air. The average was found to be 2.3 seconds.

Independent of Mass

Legend has it that Galileo Galilei performed an experiment at the Leaning Tower of Pisa. He dropped objects of different masses to see which of them falls first. Starting from the heaviest to the lightest object, what he found was that, in a vacuum, they reach the ground at the same length of time when dropped from the same height. Using this result, we can compute for the distance travelled by any object under the influence of gravity. At heights near the earth’s surface, the acceleration of objects in the presence of gravity alone is given by

$\displaystyle \frac{d^2h}{dt^2}= \frac{dv}{dt} = -g$

where $h$ is the vertical distance (height), $v$ is the velocity and $g$ is the acceleration due to gravity given by $g = 9.8 \text{m/s}^2$.

The equation above is a simple differential equation which we can solve for the velocity at any time $t$ by integrating both sides:

$\displaystyle \int^{v}_{v_i} dv = \int^{t}_{t_i} -g dt$

which by the Fundamental Theorem of Calculus gives us

$\displaystyle v - v_i = -g(t - t_i)$
$\displaystyle v = -gt + v_i + t_i$

If we take initial time of the rocket launch to be zero, that is, $t_i=0$, we have

$\displaystyle v = \frac{dh}{dt}= -gt + v_i$

We don’t know the initial velocity $v_i$ of the rocket. However, at the maximum height attained by the rocket, the velocity is zero. Using this fact, we can solve for the initial velocity:

$\displaystyle 0 = -gt_{\text{max}} + v_i$
$\displaystyle v_i = gt_{\text{max}}$

where $t_{\text{max}}$ is the time the rocket reached maximum height.

At $t_{\text{max}}=2.3$ seconds, the initial velocity is therefore $v_i = 9.8 * 2.3 = 22.5$ m/s.

Maximum height

We can solve for the height as a function of time by integrating both sides of the velocity equation:

$\displaystyle \int^{h}_{h_i} dh = \int^{t}_{t_i} -gt dt + v_i dt$
$\displaystyle h - h_i = -\frac{gt^2}{2} + v_i t \big|^t_{t_i}$

Since the initial height is zero , that is , $h_i = 0$ and the initial time is zero $t_i = 0$, we have

$\displaystyle h = -\frac{gt^2}{2} + v_i t$

Since at t=2.3 seconds, the rocket reaches the maximum height, we have

$\displaystyle h_{\text{max}} = -\frac{9.8 \times (2.3)^2}{2} + 22.5 * 2.3 = 25.829 \text{ meters}$

Does this make sense? The average height of a building storey is about 3 meters. This is about 8 or 9 storeys high which I think makes sense.

It’s good that my new year not only started with a warm welcome from friends but it also appealed to my physics curiosity.

Thank you to Marijo Condes and family for letting us spend a memorable New Year’s Eve.