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Projectile homework problem

  1. Feb 12, 2007 #1

    1. The problem statement, all variables and given/known data

    There is a link to the problem and its picture here: http://media.wiley.com/product_data/excerpt/19/04717580/0471758019-1.pdf
    It is # 46.

    Here is the problem as well:
    In Fig. 4-44, a ball is thrown up onto a roof, landing 4.00 s later at height h=20.0m above the release level. The ball’s path just before landing is angled at θ = 60.0° with the roof. (a) Find the horizontal distance d it travels. (See the hint to Problem 41.) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball’s initial velocity?


    2. Relevant equations
    I used these:
    Voy=(Y-Yo + 0.5gt²)/t
    Vo=Voy/(sin θ)
    Vox=Vo (cos θ)
    Δx=(Vo cos θ) t
    magnitude= √[(Vox)²+(Voy)²]
    θ=arctan (Y/X)

    3. The attempt at a solution
    Solved for Voy, first:
    Voy= [20 m + (0.5)(9.8 m/s²)(4 s)²]/ 4 s
    =24.6 m/s

    then, Vo:
    Vo=(24.6 m/s) / (sin 60°)=28.4 m/s

    Vox=(28.4 m/s) (cos 60°)= 14.2 m/s

    a) Δx = (28.4 m/s)(cos 60°) (4 s)
    = 56.8 m

    b) magnitude (V)= √[(14.2 m/s)²+(24.6 m/s)²]= 28.4 m/s

    c) θ= (tan ^-1)[(24.6 m/s)/(14.2 m/s)] = 60°

    I'm worried my answers seem wrong. Any help would be greatly appreciated.:biggrin:
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 13, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Since you do not know how high the ball went, you need to break the problem into 2 pieces. One piece is when the ball reaches the height of the roof (20m) reaches its highest point and returns to the roof. The second piece is the flight from the ground to roof height. You need to find how long the ball spends on the trip up to the roof. Once you get that time you know how long the rest of the flight took.

    See if you can use that to finish the problem.
  4. Feb 13, 2007 #3


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    Homework Helper

    ..... then, Vo:
    Vo=(24.6 m/s) / (sin 60°)=28.4 m/s ....

    How do you now that the launching angle is sixty degrees?

    (a) Using

    [tex]v_y = v_{y0} - gt[/tex]

    one get that

    [tex]v_{y0} = v_y + gt[/tex]

    substitute that into

    [tex]y = v_{y0}t - 0.5gt^2[/tex]

    to get

    [tex]y = v_yt + 0.5gt^2[/tex]

    using the data when it lands on the roof you can solve for the y-velocity component, which enables you to get the x-velocity component since

    [tex]\frac{v_{y\ roof}}{v_x} = \tan(60^o)[/tex]

    Do you have the entry page (contents/chapters) address for the Wiley notes?
    Last edited: Feb 13, 2007
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