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 of a Baseball Problem

  1. Feb 7, 2010 #1
    1.
    A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.5 s after being hit. Then 3.0 s after reaching its maximum height, the ball barely clears a fence that is 95.8 m from where it was hit. Assume the ground is level. What maximum height above ground level is reached by the ball? How high is the fence? How far beyond the fence does the ball strike the ground?





    2. X-Xo=(Vocosx0)t
    Y-Yo=V0yt-1/2gt^2
    V^2y=(Vosinxo)^2-2g(Y-Yo)





    3.
     
  2. jcsd
  3. Feb 8, 2010 #2
    where is your attempt to solve the question?????????????
     
  4. Feb 8, 2010 #3
    No idea where to even begin. None of my numbers are making any sense when I use these equations.
     
  5. Feb 8, 2010 #4
    First, have a look at the figure I made:

    http://img6.imageshack.us/img6/8106/14819284.jpg [Broken]

    for the first part .. it is asking for ymax , lets think at ymax what happens to the final velocity? you are given yo=0 and t at max. height =3.5s .. can you go from there?


    for the second part .. you are asked to find the height of the fence .. from the figure you can note that this height is the same height if you are aked to find y after 3s ..


    for the third part .. you can notice from the figure that if you found the range of the travel (max. x the ball travels) then finding the difference between it and 95.8m , this way you find x .. but when it hits the ground what can you say about y??


    I believe I gave you enough hints , please work or atleast try as much as you can and then show your detailed solution so if you got stuck we can help .. REMEMBER : members dont do your homework they just help :) ..
     
    Last edited by a moderator: May 4, 2017
  6. Feb 8, 2010 #5
    Ok so I found Ymax by taking my Voy^2/2g. I then found the fences height by V^2=Voy^2+2gDeltaY. I am however, stuck on the last part, I know that Delta X is going to be the total distance minus the distance given in the problem of 95.8m. But in order to find total distance I need t total but when I calculate ttotal=Vfy-Voy/Ay I get an answer of 3.5s which would mean the ball didn't make it to the fence which is incorrect?
     
  7. Feb 9, 2010 #6
    looking at the figure you can notice that there is some kind of symmetry, it should be obvious that when it reaches ymax. At t=3.5s it covers half of the distance of travel .. So from there can you find what should be the total time of the travel ??
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Projectile Motion of a Baseball Problem
Loading...