Skip to content

Time and Speeds from a Given Height

September 6, 2012

I’ve been thinking a lot about how to think about problems where you drop a ball from a certain height (let’s say its 45 m) and we want to know how long it spends in the air and how what speed it has.

The two big ideas I like to go to are

Δx = Vavg Δt

and

a = Δv /Δt

For the drop situation, assuming g = 10 m/s/s, these big ideas become

10m = vf/ 2 * t    (The average velocity is halfway between the final and the initial, which is zero)

and

10 m/s/s = vf /t   (The velocity change is vf -0)

So to solve this equation, we are looking for two numbers (and time and a velocity), and those numbers must multiply to make 20, and divide to make 10. (Or plot them and ask where does this line intersect with this hyperbola?)

My first strategy is guess and tweak. I decided since the numbers need to divide to make ten, I can easily satisfy that by picking a pair of numbers where you the second number is simply a decimal shift of the first one. For example, I picked

10 * 1 = 10 (too low)

20*2 = 40 (too high)

15 * 1.5 = 22.5 (still too high)

14 * 1.4 = 19.6 .. that’s pretty good

The ball will take about 1.4s and end up with a speed of about 14 m/s.

 

Advertisements
No comments yet

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: