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h = v0*t + 0.5*g*t^2
where
Some physics textbooks also list the following equations as being important .
v = v0 + g*t
v^2 = v0^2 + 2*g*h
where v is the velocity of the object and the other terms are the same as described above .
Let's do an exercise using the first of the two equations given above .
An individual on the surface of the earth shoots an arrow directly upward with a velocity of 100 feet per second. How many seconds elapse before the arrow turns and starts falling towards thesurface of the earth. Ignore the effects of air resistance.
Create a script
Please copy the code from Listing 6 into an html file and open the file in your browser.
Listing 6 . Exercise to find the velocity. |
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<!---------------- File JavaScript06.html ---------------------><html><body><script language="JavaScript1.3">document.write("Start Script</br></br>");
//Initialize the problem parameters.var g = -32.2;//gravity in ft/sec*sec on Earth
var v0 = 100;//initial velocity in ft/sec//Given that v = v0 + g * t
//At what time does the velocity go to zero?//Rearrange the terms in the equation.
var t = -v0/g;//Display the results
document.write("Arrow has zero velocity at " +t.toFixed(2) + " seconds " + "</br>");
document.write("</br>End Script");</script></body></html> |
Screen output
The text shown in Figure 14 should appear in your browser window when you open the html file in your browser.
Figure 14 . Screen output for Listing #6. |
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Start Script
Arrow has zero velocity at 3.11 secondsEnd Script |
Analysis of the code
Compared to working through the solutions to the previous exercises, this one seems almost trivial.
After establishing values for the acceleration of gravity and the initial velocity of the arrow, the code in Listing 6 rearranges the first equation given above and solves for the value of time at which the velocity goes to zero. This is the point in time when the arrow turnsfrom moving up and begins falling back toward the earth.
The results are shown in Figure 14 . You should compare this result with Figure 1 , which shows that the arrow reaches its maximum height at approximately 3 seconds, which agrees very well with the result shown in Figure 14 .
Let's do an exercise using the second of the two equations given above .
An individual that is six feet tall standing on the surface of the earth shoots an arrow directly upward with a velocity of 100 feet per second. What is the maximum height achieved by the arrow before it turns and falls backtowards the surface of the earth? Ignore the effects of air resistance.
Create a script
Please copy the code from Listing 7 into an html file and open the file in your browser.
Listing 7 . Exercise to find the height. |
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<!---------------- File JavaScript07.html ---------------------><html><body><script language="JavaScript1.3">document.write("Start Script</br></br>");
//Initialize the problem parameters.var g = -32.2;//gravity in ft/sec*sec on Earth
var v0 = 100;//initial velocity in ft/secvar h0 = 6;//initial height
//Given that v^2 = v0^2 + 2*g*h//What is the maximum height reached by the arrow?
//Note that the maximum height is six feet more than// the value given by the above equation because that
// equation is based on the point of release.//The maximum height occurs when the velocity goes to zero.
//Setting the velocity to zero and rearranging the terms// in the equation gives:
var h = h0 + (-(v0 * v0))/(2*g);//Display the results
document.write("Arrow reaches maximum height of " +h.toFixed(2) + " feet " + "</br>");
document.write("</br>End Script");</script></body></html> |
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