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Problem involving a projectile

  1. Sep 5, 2006 #1
    A place kicker attempts a field goal, giving the ball an initial velocity of 30 m/s at an angle of 32 degrees with the level of the field. THe uprights are 35 m from the point at which the ball is kicked and the horizontal bar is 4.0m from the ground. a)At what time after the kick will the ball pass the goal posts? B) is the kick succesful, and by how many meters does the ball clear or pass beneath the bar.

    a) v0x=30 m/s cos(32) and v0y=30 m/s sin(32)
    v0x=25.44 m/s m/s and v0y=15.89 m/s

    t= 35/25.44 seconds

    b) how do i aproach this lol?
    Last edited: Sep 5, 2006
  2. jcsd
  3. Sep 5, 2006 #2
    t = time elapsed after the kick till the ball pass the goal posts = 1.38 s
    y(t) = v0yt - 0.5gt2

    Did it help?
  4. Sep 5, 2006 #3
    First off I think your time is wrong, you said 40m/s to start with, but changed to 30.

    To do b), you need to find y(t), then plug in the t you got from the first part. To do this, integrate up F=md2y/dt2=mg twice, remembering the initial conditions.
  5. Sep 5, 2006 #4
    15.89 * (1.38) - 0.5(9.8^2)(1.38^2)?

    then would i subtract it by 5?
  6. Sep 5, 2006 #5
    is that right?
  7. Sep 6, 2006 #6
    close. it would be 15.89 m/s * 1.38 s - 0.5 (9.8 m/s^2)(1.38 s)^2 [note that the 9.8 is not squared] and then compare this answer to the height of the bar.
  8. Sep 6, 2006 #7
    Thanks for the help man, wont be long till i come back!
  9. Sep 6, 2006 #8
    ohk now..this one's easy...all one needs 2 understand is that we need 2 compare d distance travelled in vertical direction exactly at d instant wen d ball reaches d goal .post with d height of the horizontal bar...
    eqn comes to s(y)= 15.89* 35/25.44 - 1/2* 9.8 *(35/25.44)^2
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