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Homework Help: Throwing a projectile so that it clears the window

  1. Jan 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Standing on the ground 3.0 m from the wall of a building, you want to throw a package from your 1.6 m shoulder level to someone in a second-floor window 4.1 m above the ground

    a)At what speed should you throw the package so it just barely reaches the window?
    b) At what angle should you throw the package so it just barely reaches the window?

    I know how to solve it conceptually but the workings are causing me problems mainly because I am asked to determine the 2 crucial unknowns.

    3. The attempt at a solution

    Assuming the projectile crosses the window at the apex of the trajectory.

    vyf = vyi -gt
    0 = vyi -9.8t
    t = vi sin Θ/g

    y(vi sin Θ/g) = y(0) +vi.t - 0.5gt^2
    y(vi sin Θ/g) - y(0) = (vi sin Θ)(vi sin Θ/g) - 4.9(vi sin Θ/g)^2
    2.7 = [(vi ^2 sin^2Θ)/g] -4.9[(vi^2 sin^Θ)/g^2]

    I cannot assume Θ= 45 because part b asked for the angle. However, without knowing the angle, I cannot determine the initial velocity-self-contradictory, if you ask me.
  2. jcsd
  3. Jan 23, 2014 #2
    It's not a very obvious question is it. I think it's looking for when it has no vertical speed, what else would barely reaches the window mean? In that case it's not to bad.
    Vertical component:
    s=2.5 u=? v=0 a=-9.81 t=? s=vt-0.5at^2, 2.5=-0.5*-9.81*t^2, t=0.714, also can find using other suvat equation u=7.

    Solve for horizontal using t=0.714 find that u=4.20

    overall magnitude=(7^2+4.2^2)^0.5=8.16.
    Angle=tan^-1(7/4.2)=59 degrees.
  4. Jan 23, 2014 #3
    You don't need to assume an angle. You just need to break it up into 2 parts initial velocity in y, and then in x.

    Initial velocity in y would the speed needed to reach 2.5m(from the shoulder). And x is a little more tricky, you need to realize that it will reach its apex half way through its journey in x, which you would want it to travel 6m before returning to the height it was thrown from.
  5. Jan 24, 2014 #4

    V=0? Is v the initial speed? If so, it doesn't make sense.
    I'll work on it tonight.
  6. Jan 24, 2014 #5
    V= 0 would be the velocity at its apex.
  7. Jan 24, 2014 #6
    The velocity in y that is.
  8. Jan 24, 2014 #7

    At the apex yes. That would be the final velocity using vyf^2 -vyi^2 =2g(yf-yi)
    But the other poster was using the equation
    Yf= yi + vyi.t -0.5gt^2 and denoting vyi =0
  9. Jan 24, 2014 #8
    Well, obviously you know that the initial velocity can't be 0. But yes, I would definitely use the formula you just posted, there is no reason to use time if you don't have to.
  10. Jan 24, 2014 #9
    0 = vyi^2 = 2(-9.8)(4.2-1.5)
    vyi = 7.27 ms^-1

    The answer to part (a) is apparently 8.3ms^-1

    What is wrong here?
  11. Jan 24, 2014 #10
    That's only the y component, now find the x component.
  12. Jan 24, 2014 #11
    Also the numbers you used 4.2 and 1.6 should be 4.1 and 1.6 according to the op.
  13. Jan 24, 2014 #12
    That explains. I've to find the y and x component and find the magnitude, right?
    I think there's another way to determine the y-component and that is to find the height of the trajectory.
    However, from sources, it says, in using the height trajectory formula, the time, t = vyi/g must be substitute into y = yaverage velocity x t. Doesn't it make more sense to sub time into y(t) = y(0) + vi.t - 0.5gt^2?
    Last edited: Jan 24, 2014
  14. Jan 24, 2014 #13
    In the same time period, t = vi sinΘ/g, the projectile moves a horizontal distance of
    x = vi.t = vi cosΘ(vi sinΘ/g), am I right?
  15. Jan 24, 2014 #14
    In the same time period, t = vi sinΘ/g, the projectile moves a horizontal distance of
    x = vi.t = vi cosΘ(vi sinΘ/g), am I right?
    It should be right but apparently, there are too many unknown variables.

    x = vi^2 sin(2Θ)/2g
    x = 3m and g = 9.8 but I have no knowledge of Θ and vi

    3 m = vi^2sin(2Θ)/2g
    vi^2 sin(2Θ) = 58.8
    Unable to proceed further.
    Last edited: Jan 24, 2014
  16. Jan 24, 2014 #15


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    You haven't used the height. I think if you write another equation using the height it will work out but I haven't tried.
  17. Jan 24, 2014 #16
    Last edited: Jan 24, 2014
  18. Jan 24, 2014 #17
    It's the same thing. They both give me the same y-component value.
    The tricky part is to solve for the x-component.
    I know vyi = 7.27.
    In order to determine the numerical time value, I sub vyi into y(t) = vi.t -0.5gt^2
    The quadratic equation is 2.7 = 7.27t - 4.9t^2
    The equation cannot be solve because the value to be square root is a negative.
  19. Jan 24, 2014 #18
    Solved. I made a arithmetic error.
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