Math Help (Physics)
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UltraMario
Donkey Kong
Off-topic, but your math problems have VIDEOGAME CHARACTERS? I wish I was in your class. 
However, this problem I am not sure how to solve.
However, this problem I am not sure how to solve.
New Super Mario
Ghirahim
so, what set up a triangle or something or a proportion?
Really, I have no idea, that's all that came to mind.
Really, I have no idea, that's all that came to mind.
I'd scan a picture of the diagram plus the hand calculations, but the scanner's on the fritz so hopefully a written explanation will suffice, plus this related diagram I found online (just ignore the values and the ymax bit: the x-intercept is what we're looking for):
Use 782 as dy (vertical height) and calculate for t using -49sin(45°) as Vy1 (courtesy of trigonometry: I assume I don't need to explain why) and 9.8 as ay (as you can see, going down is positive and up is negative: it's simpler to work with):
dy = vy1*t + 0.5*ay*t2
782 = -49sin(45°)*t + 0.5*(9.8)*t2
782 = -34.65*t + 4.9*t2
0 = 4.9*t2 - 34.65*t - 782
From there use the quadratic formula (I assume you know this too) to get t (reject the negative value):
t = 16.65
Then plug that into the dx calulation, using 49cos(45°) as Vx (trig again):
dx = vx*t
dx = 49cos(45°)*(16.65)
dx = 34.65*16.65
dx = 576.92
Therefore, Mega Man travels approximately 577 km.
(although gravity is actually 9.8 m/s, not km/s, and all the heights and speeds should be in m too)
Use 782 as dy (vertical height) and calculate for t using -49sin(45°) as Vy1 (courtesy of trigonometry: I assume I don't need to explain why) and 9.8 as ay (as you can see, going down is positive and up is negative: it's simpler to work with):
dy = vy1*t + 0.5*ay*t2
782 = -49sin(45°)*t + 0.5*(9.8)*t2
782 = -34.65*t + 4.9*t2
0 = 4.9*t2 - 34.65*t - 782
From there use the quadratic formula (I assume you know this too) to get t (reject the negative value):
t = 16.65
Then plug that into the dx calulation, using 49cos(45°) as Vx (trig again):
dx = vx*t
dx = 49cos(45°)*(16.65)
dx = 34.65*16.65
dx = 576.92
Therefore, Mega Man travels approximately 577 km.
(although gravity is actually 9.8 m/s, not km/s, and all the heights and speeds should be in m too)
My high school physics teacher always used Snoopy and Woodstock for our word problems.Baby Yoshi said:Off-topic, but your math problems have VIDEOGAME CHARACTERS? I wish I was in your class.
GalacticPetey
Donkey Kong
My math teacher once used Superman and Lex Luthor.
GalacticPetey said:
He's special.
ralphfan
Thank you based god
So we're talking horizontal distance here?
Not sure what to do about the angle, though.
EDIT:
If we're just trying to calculate horizontal distance, you may be able to do it this way:
First, calculate distance, velocity, acceleration and time for the vertical distance. We know he's falling 782 meters. We also know he's accelerating at 9.8 km/s/s. Initial velocity is naturally, for free fall, 0 km/s. Using the equation d=1/2at2, we can find that t equals 12.63. We then need to know how far he traveled horizontally. We do so using the information we just found.
We know that he's going at 49 km/s and will be in the air for 12.63 seconds. That would mean he covered 618.87 km in this time.
However, this problem may require the more complex calculations that Walkazo used because of the jump at a 45-degree angle.
I'd like to know what the correct answer is. I'd assume Walkazo is right here.
Not sure what to do about the angle, though.
EDIT:
If we're just trying to calculate horizontal distance, you may be able to do it this way:
First, calculate distance, velocity, acceleration and time for the vertical distance. We know he's falling 782 meters. We also know he's accelerating at 9.8 km/s/s. Initial velocity is naturally, for free fall, 0 km/s. Using the equation d=1/2at2, we can find that t equals 12.63. We then need to know how far he traveled horizontally. We do so using the information we just found.
We know that he's going at 49 km/s and will be in the air for 12.63 seconds. That would mean he covered 618.87 km in this time.
However, this problem may require the more complex calculations that Walkazo used because of the jump at a 45-degree angle.
I'd like to know what the correct answer is. I'd assume Walkazo is right here.
UltraMario
Donkey Kong
If only Pi were here. 
This was a standard question back in my Grade 12 physics class: I even looked at my old notes to see exactly how it's supposed to be answered. Of course, different schools might use different strategies, but that's what I was taught and it should work.Ralph said:
Initial velocity isn't 0, it's 49 km/h. You use trigonometry to break that into horizontal and vertical components:
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