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!

Homework Help: Projectile Motion Problem with thrown ball off roof

  1. Jun 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown leftward from the left edge of the roof, at height [tex]h[/tex] above the ground. The ball hits

    the ground [tex]1.50s[/tex] later, at distance [tex]d=25.0m[/tex] from the building and at angle [tex]\theta =60^

    {\circ}[/tex] with the horizontal. (a) Find h. What are the (b) Magnitude and (c) angle relative to the horizontal

    of the velocity at which the ball is thrown? (d) Is the angle above or below the horizontal?

    http://lcphr3ak.is-a-geek.com/fig49.png [Broken]


    The answers in the back are:

    2. Relevant equations
    [tex]x - x_{\circ} = v_{\circ} t - \frac{1}{2} a t^2[/tex]
    [tex]v_{\circ x} = v_{\circ} \cos{\theta _{\circ}}[/tex]
    [tex]v_{\circ y} = v_{\circ} \sin{\theta _{\circ}}[/tex]

    3. The attempt at a solution

    I'm missing an observation here, because I don't see enough information to solve this. This problem is suppose to

    be solved with projectile motion. So first I attempt to find [tex]v_{fx}[/tex] (which is the [tex]x[/tex] component

    of the vector at [tex]\theta[/tex]): [tex]x - x_{\circ} = v_{fx} t - \frac{1}{2} a t^2[/tex] which is [tex]-25 =

    v_{fx} (1.50)[/tex]. This gives me [tex]v_{fx} = -16.667[/tex]. Using this, I take [tex]v_{fx} = v_f \cos{\theta}

    [/tex], which is: [tex]-16.667 = v_f \cos{60}[/tex], so [tex]v_f = -33.32[/tex]. Using this, I find [tex]v_{fy}

    [/tex] by taking [tex]v_{fy} = v_f \sin{\theta}[/tex] and doing [tex]v_{fy} = (-33.32)\sin{60}[/tex] so [tex]v_{fy}

    = -28.856[/tex]. So, to find [tex]h[/tex], we solve for [tex]y[/tex]. [tex]y = v_{fy} t - \frac{1}{2} a t^2[/tex]

    becomes [tex]y = (-28.856)(1.50) - \frac{1}{2} (9.8) (1.50)^2[/tex] which says [tex]y = -54.309[/tex]. Now, the

    absolute value of [tex]y[/tex] should be the height, but this is waaaaay off from the answer. I'm practically out of


    Could anyone explain what I'm doing wrong? If I didn't provide enough info, just let me know and I'll try to

    provide more.

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 6, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi coco87! :smile:

    (have a theta: θ :wink:)
    Nooo … θ in your formula is the angle at the roof. not the θ (= 60º) in the diagram! :smile:
  4. Jun 6, 2009 #3
    Take care with the sign of the acceleration.
    It is often helpful to write down on your diagram which direction you are taking as positive.
  5. Jun 6, 2009 #4
    So, I'm either using the wrong formula, or I need to find the initial velocity (or angle)? I don't know of any other formulas (other than [tex]v = v_{\circ}+at[/tex]), and I have no idea how I would even obtain the angle up there. Could you possibly give me a hint? (or another one :wink: )

    Thanks for the responce!
  6. Jun 6, 2009 #5
    you can use that formula to find the initial vertical velocity. Then you can work out an angle!
  7. Jun 6, 2009 #6
    What you had said triggered an idea, and after playing around with the equations, I found out that I had misunderstood the Constant Accelloration equations all along.. well, I do now :biggrin:

    Thank you guys for your help!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook