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Homework Help: Projectile motion Problem(Is My Ans Correct?)

  1. Mar 8, 2012 #1
    The question: An object has a Velocity V1 and an angle A (Alpha) initially. Later in its path, this same object has an Angle B (Beta)

    find the time elapsed from point A to point B

    Okay so i have an answer.. i just need some one to tell me if what i'm doing is correct./answer is correct.

    http://postimage.org/image/xsei89eg1/ [Broken]

    here is an image of the problem.

    Okay so..

    V1x = V1cosA
    V1y = V1sinA

    Let V2 be the variable for the second velocity.

    Since horizontal component of velocity is constant through projectile motion..


    V2x = v1cosA

    Through trigonemtry we can resolve the Y component for V2, which i need for further calculation

    TanB = V2y/V2x

    V2y= V2xTanB

    Now i have all i need and i can use uniform accelerated formula, to solve for The time...

    Vf=Vi + gt

    V2y = V1y + gt

    (v2y - v1y)/g = t

    Answer: t = (v1cosATanB - V1sinA)/g

    or simplifieed too... V1(cosA*TanB - SinA)/g


    = s (seconds)

    Units match.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 8, 2012 #2


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    Homework Helper

    Got the same thing with opposite sign - I took g to be 9.81; in your solution g = -9.81.
  4. Mar 9, 2012 #3
    lol i always make taht stupid mistake.. xD.

    so t= (v1cosAtanB - v1sina) /-g

    t = (v1sina - v1cosAtanB )/ g
  5. Mar 11, 2012 #4

    so using your frame of reference for the vertical component as g = 9.81 and X portion being positive in the direction of motion

    TanB = -V2y/v2x

    -V2y = TanB*V1cosA
    v2y = -TanB*v1cosA

    v2y = v1y + gt

    -TanB*v1CosA = -V1SinA + Gt
    T = (V1SinA - V1CosA*TanB)/g

    T= V1(SinA-CosA*TanB)/g

    I'm just curious for what i did in the post above, am i 'allowed' to do that..

    cause i thoguht -a/b is the same thing as a/-b

    ? Also i prefer using your way especially when using purely variables making;
    Gravity = Positive Direction

    because i don't have to account for the sign, therefore i wont lose track of it.
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