- #1
Lostone
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First off, I'm glad I stumbled upon this place. Secondly, taking physics as an online course is probably not the best judgment call I have made in recent times. However, I'm here now and I plan on sticking through with the course. My problems, however, are lying with the labs. Since they give us websites and a book, I'm attempting at piecing things together as best I can. The website they gave us for this particular lab assignment is here: http://www.hazelwood.k12.mo.us/~grichert/explore/dswmedia/range.htm"
This is regarding a 'no air' set up. The initial setting for the course is V= 50m/s and a launch angle of 45degrees. The time it takes for the shot to land is 7.1s. (though there is a part where the air will need to be turned 'on'. According to the site, it's based on vc, velocity and a constant *which I would assume to be gravity*)
The lab wants the horizontal displacement, which, after having to poke the internet for some clues, I believe equates to D= Vi t + 1/2at^2. I've also seen something much simpler looking at d=vt.
Utilizing the first equation,
D=(50m/s)(7.1s)+ 1/2(0)(7.1s) *7.1 is the total time so no squaring...I think*
D=(355)(m/s)(s/1)+(0)
D=355m
Utilizing the second,
d=(50m/s)(7.1s)
d=355(m/s)(s/1)
d=355m
Now, here's the catch. According to the graph supplied with the website, I'm showing an answer that is 100m too far as it lands well before the 300m mark. That and I don't see where the 45 degree angle came into play on the equation itself.
This is just the starter question. There are various angles in which a displacement is needed but I don't understand how to get the starter one up first. Any help, please?
Homework Statement
This is regarding a 'no air' set up. The initial setting for the course is V= 50m/s and a launch angle of 45degrees. The time it takes for the shot to land is 7.1s. (though there is a part where the air will need to be turned 'on'. According to the site, it's based on vc, velocity and a constant *which I would assume to be gravity*)
Homework Equations
The lab wants the horizontal displacement, which, after having to poke the internet for some clues, I believe equates to D= Vi t + 1/2at^2. I've also seen something much simpler looking at d=vt.
The Attempt at a Solution
Utilizing the first equation,
D=(50m/s)(7.1s)+ 1/2(0)(7.1s) *7.1 is the total time so no squaring...I think*
D=(355)(m/s)(s/1)+(0)
D=355m
Utilizing the second,
d=(50m/s)(7.1s)
d=355(m/s)(s/1)
d=355m
Now, here's the catch. According to the graph supplied with the website, I'm showing an answer that is 100m too far as it lands well before the 300m mark. That and I don't see where the 45 degree angle came into play on the equation itself.
This is just the starter question. There are various angles in which a displacement is needed but I don't understand how to get the starter one up first. Any help, please?
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