1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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..

    V1x=V2x

    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

    (m/s)/(m/s^2)

    = s (seconds)

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

    Delphi51

    User Avatar
    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
    d

    so using your frame of reference for the vertical component as g = 9.81 and X portion being positive in the direction of motion
    v2x=v1x

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook