1. Not finding help here? Sign up for a free 30min 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!

Finding y in a projectile motion problem

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data
    A golfer hits a shot to a green that is elevated 4.0 m above the point where the ball is struck. The ball leaves the club at a speed of 15.0 m/s at an angle of 40.0° above the horizontal, which is the +x axis. It rises in the +y direction to its maximum height and then falls down to the green. Ignore air resistance.

    What is the algebraic expression for the y component vy of the ball's velocity just before landing on the green? Calculate for the y component vy using this equation.

    a_x= 0
    a_y= 9.8 m/s^2
    v_x (initial) = cos40*15 = +11.5 m/s

    2. Relevant equations

    i thought it would be v^2=v(initial)^2 + 2ay rearranged to solve for just v(y direction). therefore: square root of v(initial)^2 + 2ay.
    y being the distance in the vertical direction.

    also: for v(initial) i took the sin40*15 to get 9.6 m/s.

    3. The attempt at a solution
    when i plug this into my homework website, it tells me i am wrong. but this is the only kinematic that i can think of using only v(initial), y displacement, and acceleration. what's wrong with my equation?
  2. jcsd
  3. May 27, 2009 #2


    User Avatar
    Homework Helper

    What value did you get? Did you use [itex]v_y^2=(v_0)^2_y+2ay[/itex] or [itex]v_y^2=(v_0)^2_y-2ay[/itex] ?
  4. May 27, 2009 #3
    i used the equation LaTeX Code: v_y^2=(v_0)^2_y-2ay and rearranged it so it would be sqtroot(v(initial)^2 + 2ay). i think i got like -8.9 m/s, but the website told me i was wrong.
  5. May 27, 2009 #4
    by the way i have to write it in a way solving for just v_y
    Last edited: May 27, 2009
  6. May 27, 2009 #5


    User Avatar
    Homework Helper


    if you use v2=u2+2ay then you are saying that acceleration is acting upwards. The only acceleration in this question is gravity which acts downwards.So, you'd need to use v2=u2-2ay
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?