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Projectile motion at an angle

  1. Dec 13, 2004 #1

    kreil

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    Gold Member

    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.

    Vvi=(sin50)(3.37m/s)=2.581m/s
    Vhi=(cos50)(3.37m/s)=2.166m/s

    [tex] vertical:[/tex]
    [tex]v_i=2.581\frac{m}{s}[/tex]
    [tex]d=1.15m[/tex]
    [tex]a=g=9.81\frac{m}{s^2}[/tex]
    [tex]t=?[/tex]

    [tex]d=v_it+\frac{1}{2}at^2[/tex]

    [tex]1.15=2.581t+4.905t^2[/tex]

    [tex]4.905t^2+2.581t-1.15=0[/tex]

    [tex]t=\frac{-2.581+\sqrt{6.661561+22.563}}{9.81}[/tex]

    [tex]t=\frac{2.82497}{9.81}=.288s[/tex]

    [tex]horizontal:[/tex]
    [tex]t=.288s[/tex]
    [tex]v_i=2.166\frac{m}{s}[/tex]
    [tex]a=0[/tex]
    [tex]d=?[/tex]

    [tex]d=v_it[/tex]
    [tex]d=(2.166)(.288)=.6238m[/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.

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

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

    Tide

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    Science Advisor
    Homework Helper

    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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