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Projectile Motion- Horizontal Range and Time of Flight

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A rifle is used to fire two bullets. The first shot is at an angle of 60° above the horizontal and the second at an angle of 45° above the horizontal. The speed of EACH bullet as it leaves the rifle is 200ms^1. For EACH bullet calculate the horizontal range and the corresponding time of flight.

    2. Relevant equations

    Horizontal Range: v^2sin2ө/ g

    Time of Flight= 2vsinө/g

    3. The attempt at a solution

    Horizontal Range for 1st shot= (200^2)sin2(60)/9.81

    =3531m

    Time of Flight= 2(200)sin(60)/9.81

    =35.3 secs

    Horizontal Range for 2nd shot= (200^2)sin2(45)/9.81

    =4077m

    Time of Flight= 2(200)sin(45)/9.81

    = 28.8 secs

    I'm not sure if I'm supposed to use the 200 as v or if I'm to calculate v somehow from the given information.:confused:
     
    Last edited: Mar 10, 2010
  2. jcsd
  3. Mar 10, 2010 #2
    You have used the equations correctly. I guess you are just not sure how the equations work? Maybe reading the proofs of these equations would help.
     
  4. Mar 10, 2010 #3
    ok...i just keep thinking that since it left the rifle at 200ms^1that that would be the initial velocity, u. I wasn't too sure if I could use it in the equation as v. I thought that i some how had to calculate it but then there would be more than one unknowns when using the equatios of motion:redface:
     
  5. Mar 10, 2010 #4
    ok...i just keep thinking that since it left the rifle at 200ms^1that that would be the initial velocity, u. I wasn't too sure if I could use it in the equation as v. I thought that i some how had to calculate it but then there would be more than one unknowns when using the equations of motion:redface:
     
  6. Mar 10, 2010 #5

    ideasrule

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    Homework Helper

    v represents initial velocity, so it should be 200 m/s.
     
  7. Mar 10, 2010 #6
    ok thank you
     
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