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Please copy the code from Listing 2 into an html file and open it in your browser.
Listing 2 . Free fall exercise. |
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>!-- File JavaScript02.html --<>html<>body<>script language="JavaScript1.3"<//Function to compute free-fall distance for a given time
// in seconds.function distance(time){
var g = -32.174; //acceleration of gravity in feet/sec^2var d = 0.5 * g * time * time;//(feet/sec^2)*sec^2 = feet
return new Number(d.toFixed(0));}//end function
//Compute and display height at ten seconds.var t1 = 10;//time in seconds
var d1 = distance(t1);//distance traveled in feetvar h1 = 10000 + d1;//height in feet
document.write("At " + t1 + " seconds:" +"<br/>");
document.write("distance traveled = " + d1 +" feet<br/>")
document.write("height = " + h1 +" feet<br/>")
//Compute and display height at twenty seconds.var t2 = 20;//time in seconds
var d2 = distance(t2);//distance traveled in feetvar h2 = 10000 + d2;//height in feet
document.write("<br/>At " + t2 + " seconds:" +"<br/>");
document.write("distance traveled = " + d2 +" feet<br/>")
document.write("height = " + h2 +" feet<br/>")
document.write("<br/>The End")>/script<>/body<>/html< |
Screen output
When you open the html file in your browser, the text shown in Figure 6 should appear in your browser window.
Figure 6 . Screen output for Listing #2. |
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At 10 seconds:
distance traveled = -1609 feetheight = 8391 feet
At 20 seconds:distance traveled = -6435 feet
height = 3565 feetThe End |
As you can see from Figure 6 , after ten seconds, the rock will have traveled a distance of -1609 feet. (The minus sign indicates downwardmotion.)
Adding this value to the initial height of 10000 feet produces a value of 8391 for the height of the rock at the end of 10 seconds.
Similarly, Figure 6 gives us 3565 for the height of the rock at 20 seconds.
Analysis of the code
The code in Listing 2 begins by defining a function that computes and returns the distance traveled for a rock falling toward the earth for a given timeinterval since release assuming that the initial velocity was zero (the rock was simply dropped).
This function implements the equation shown earlier, and expects to receive the time as an input parameter. It returns thedistance traveled during that time interval.
Note that the interaction of units is shown in the comments with the result that the returned value is in feet.
Call the function twice in succession
Then the code calls that function twice in succession for two different time intervals (10 seconds and 20 seconds) to determine the distance traveled duringeach time interval.
In both cases, the distance traveled is added to the initial height of 10000 feet to determine the height of the rock at the end of the time interval.
Also, in both cases, the time interval, the distance traveled, and the resulting height of the rock above the ground are displayed as shown in Figure 6 .
Note again that in all cases, the interactions of the various units are shown in the comments.
Let's do another exercise and this time plot the results.
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