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Homework Help: Project motion! help please.

  1. Mar 6, 2010 #1
    1. The problem statement, all variables and given/known data

    a ball is projected at an angle of 37degree with respect to the horizontal with an initial speed of 25ft/s. if it is traveling horizontally when it hits a verical wall,
    what is the height of the wall?

    2. Relevant equations

    3. The attempt at a solution

    a) Vxi = 25 x cos37 = approximately 20ft/s , Vyi = 15 x sin37 = app 15ft/s
    height of wall = Y axis displacement
    so delta Y = Vyi x delta time + 1/2 x gravity x delta time^2
    there is no information about displacement to X axis
    in this case how can I get delta time?
  2. jcsd
  3. Mar 6, 2010 #2


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    The solution is using the vertical velocity.
    First of all, what is the formula for it?
    Secondly, how can you tell when the ball is moving horizontally?
  4. Mar 6, 2010 #3
    hm.. why do we have to use the vertical velocity?
    so whicn mean Vy= Vyi - gt ? that what should I do?
    since there is no displacement to X axis, I am not able to find delta time.
    I am wondering if you understood this problem. since I wrote whole question from the book
    thank you !
  5. Mar 6, 2010 #4


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    I understood the question alright, but did you?

    You said:
    I claimed that you can convert this into an equation, and asked you if you have any clue which equation that is. (Hint: it has to do with the vertical velocity). From this equation, you will be able to solve the delta time that you need for the horizontal distance.
  6. Mar 6, 2010 #5

    if it is travelling horizontally = there is no vertical velocity.
    that is what I know. am I right?

    vertical V equation is Vy = Vyi + gravity x delta times
    so 0 = Vi x sin37 + (-32) x delta times
    = approximately 15 ft/s^2 + (-32) x delta times
    so that delta times is 15/32 sec
    and the height of the wall = displacement to y axis
    so Y = Yo + Vyi(=15) x delta times + 1/2 x G x delta times^2
    = 0 + 15 x 15/32 + (-16) x (15/32)^2
    = 0 + 7 + (-3.5)
    so height is approximately 3.5 ft

    do u agree with me?

    and I want you to confirm one more question followed by this problem.
    " if when the object bounces away from the wall, its speed is reduced to one half of its original speed, then how far away from the wall does it hit the ground ? "
    this is second question.

    I've tried to resolve this problem as well.
    here is my attemption.

    since height of wall is 3.5 ft
    delta Y = -3.5 = Vyi - g x delta times
    = Vyi(=0) -32 x delta times
    so delta times is approximately 0.1 sec ( in this case, initial velocity of Y axia is 0. right? )
    btw i want to know x displacement. so X = Xo + Vix x delta times + 1/2 x acceleration x delta times^2
    = 0 + 10 x delta times + 1/2 x acceleration(=0) x delta times^2
    = 0 + 10 x 0.1 + 0 = 1 ft
    so distance from the wall is 1ft
    Vix = 25 x cos37 / 2 ( since its speed is reduced ) = app 10ft/s

    Am i doing right? anyway thank you for the willingness to help me out :)
    Last edited: Mar 6, 2010
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