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!

Simple Projectile Motion Problem

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Baseball is hit at 60.0m/s at an angle of 30 degrees to the horizontal. Assuming the field is level find:
    Time ball was in air
    How long it was in the air
    How far did it go
    Given
    Assume up is +ve
    Calculated data within brackets
    Horizontal
    v=(52m/s)
    d=
    t=
    Vertical
    vi=(30m/s)
    vf=
    a=-9.8m/s
    d=
    t=
    Ball
    v=60.0m/s[Forward30Up]
    2. Relevant equations
    sinθ=opp/hyp
    cosθ=adj/hyp
    d=ut+0.5at^2
    v=d/t

    3. The attempt at a solution
    Assume (variable)(h) means a horizontal variable and (variable)(v) means the vertical variable.
    So i drew out the vector diagram for the ball, with v(ball)=v(h)+v(v), knowing angle is 30
    Using sine and cosine i was able to determine the values for v(h)=52m/s and v(v)=30m/s.
    (Calculated info written in the givens within brackets)
    So now I am stuck...I am not given enough information to determine any other variables, I think?
     
  2. jcsd
  3. Feb 21, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi PinguNinja! Welcome to PF! :wink:
    Yes, that's fine so far. :smile:

    Now use the standard constant acceleration equation in the vertical direction, to find t.

    (the displacement d = 0 :wink:)​
     
  4. Feb 21, 2012 #3
    Perhaps this is a silly question, but why would the displacement be zero? Were the displacement be zero, would that not mean that the ball is at the start position? (Start position being where the ball is hit). If we were to set the point it hit the ground to be zero, then would we not run into the problem where we do not know at what height the start position should be?

    d=ut+0.5at^2
    0=30t+0.5(-9.8)t^2
    -30t=0.5(-9.8)t^2
    6.1=t
     
  5. Feb 21, 2012 #4
    Assuming the ball is initially on the ground (call this h = 0) and it is hit with a bat at an angle of 30 degrees above the horizontal we can say that the displacement is 0 becasue the ball travels up in the air to its max height (call this h = [itex]h_{max}[/itex]; it then falls to the ground returning to h = 0 - so the displacement of the ball is 0 (it returned to the height from which it was hit).

    If you were to hold the ball above the ground at some arbitrary height (say, h = [itex]h_{o}[/itex] it would rise to its max height [itex]h_{max}[/itex], then fall to its original height [itex]h_{o}[/itex] and continue falling until it hit the ground. In this case the total displacement would be [itex] - h_{o}[/itex] because the change in its position would be [itex]\Delta h = h_{final} - h_{initial} = 0 - h_{o}[/itex].
     
  6. Feb 21, 2012 #5
    Alright. So since it appears that the question is not doable without assumptions (i assume? :wink:) Then I will assume that the ball hits the bat when the bat is touching the ground :3.
     
  7. Feb 21, 2012 #6
    That seems like the best assumption to make, in my opinion. :)
     
  8. Feb 22, 2012 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi PinguNinja! :smile:

    (just got up :zzz:)

    Yes, I agree with Tsunoyukami :smile: … the question is badly worded. :frown:

    However, you can work it out from …
    clearly this was put there for some purpose,

    and the writer obviously thought he was saying that the vertical displacement was zero! :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple Projectile Motion Problem
Loading...