Mary
Project Title
Vertical Marble
Author
Problem and solution by Mary Dunlop '98
Updated 05/20/99
IP File and formatting by Peter Rapp '99
Issue:
    A marble is shot vertically upwards at 45 m/s. How high is it above the ground after 6.6 seconds?
    This is one of the most elementary physics problems, one that most classes will get to right at the beginning of the year. The calculations are pretty much straightforward. Gravity can be defined as 10m/s^2 to simplify things. In "real life" gravity is closer to 9.81, but here it is more important just to get the concept, rather than the exact answer.


Solution: 
    The QuickTime movie on the left shows the motion of the ball over time. Use the slider below the movie, and step to the frame where time is 6.6 seconds. If your answer is correct, good job! Otherwise, you may want to try working out the problem again before peeking at the equations below.


Vo=45m/s; s=?m; t=6.6sec; a=-10m/s^2
Here is what we are given.
We'll use -10 for gravity just to simplify calculations.

s= Vot+.5at^2
Memorize this relationship. It will be useful no matter what level of physics you are in. Remember Vo is initial velocity.

s=(45)(6.6)+(1/2)(-10)(6.6)^2
Here, we plug in the values that we are given.
This is simple substitution.


Resources Window
Interactive Physics is a powerful simulation program that is available from MSC Working Knowledge. IP documents are playable with Interactive Physics v1.2 or later. Earlier versions may be compatable.
Text versions of the equation work are made with Microsoft Equation Editor and are embedded in Microsoft Word documents.
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