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Homework Help: Projectile motion at an angle

  1. Dec 13, 2004 #1


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    A slug is shot at a 50 degree angle from a launcher 1.15m off the ground and has a muzzle velocity of 3.37m/s. Find the range of the slug.

    I solved it this way but I'm not sure if I did it correctly.


    [tex] vertical:[/tex]








    so my answer is .6238m...but this doesn't seem correct at all. During the lab the slug went an average of 1.4 m each trial. Someone help me!
  2. jcsd
  3. Dec 13, 2004 #2
    The reason you are getting wrong anwsers is because of the value you are using for the d. The formula you are thinking of is y(t)= y(intitial) + V(intial)t - 1/2 gt^2. By inputting 1.15 as d you are saying that the max height is going to be 1.15m. This is incorect. To find the max height H= (V(initial))^2(sinx)^2 / 2g (I derived this from the formula V^2 = V(inti)^2 + 2a(y-y(intial). Then plgu in the value from the height into the intial equation and solve for time. After soving for time use the standard d=vt.

    Use the formula
    Range= (V(intial)^2)(sin2x) / g

    Hopefully I have been helpful.
  4. Dec 13, 2004 #3


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    You have the wrong equation for the height. The acceleration is downward so you want a = -g. With your equation the projectile will continue to rise forever - I don't think that's consistent with your observations. :-)
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