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Ball from the ground lands on top of a building

  1. May 2, 2013 #1
    Well, I took my Physics final today, and I passed it but I didn't do quite as well as I'd hoped. Anyway, there was one problem I spent tons of time trying to figure out. I've tried giving it a shot again at home, but it's still eluding me. Some guidance would be great! Sorry if this is vague, but I'm writing this from memory.

    1. The problem statement, all variables and given/known data
    A ball is thrown from the ground and after 4 seconds lands on top of a building 20m high. When it lands, it is at an angle of 60°. Find the distance traveled and the angle from the ground with respect to the x-axis.

    Known:
    Landing Angle, θ=60°
    Time = 4s
    Height of building = 20m

    And my crude drawing is attached.


    2. Relevant equations

    KE and PE
    Kinematics

    3. The attempt at a solution

    I'm not trying to solve at this point, I just want to be able to find the first unknown.

    So I've tried everything I can think of to try. I was leaning toward using PE at the top of the curve to equal KE right before it hits. I'll call H the max height and h the height of the building. This would leave me with mgH = 1/2mv^2*sin(θ) + mgh
    From here, I have two unknowns. The final velocity and the maximum height, and it just seems like no matter what other equations I use, I still have two unknowns. Can someone give me a hint?
     

    Attached Files:

  2. jcsd
  3. May 2, 2013 #2
    I'm going to be a tad upset if I just figured this out:

    Using kinematics from the top of the parabola:
    for the Y
    Vf=Vi + at
    Vf= 0 + 10*4
    Vf= 40m/s

    Plug into equation from my first post:

    mgH= 1/2*mv^2 + mgh
    H= (1/2*v^2 + gh)/g
    H= 100m

    Edit:
    No, this isn't right either! It's not 4 seconds from the top of the parabola to the roof!
     
    Last edited: May 2, 2013
  4. May 2, 2013 #3

    jedishrfu

    Staff: Mentor

    Dont feel bad one of my friends and I took an EM exam once and he insisted on talking about one problem spouting out the answer in x, y, and z until I said it was a 2D problem and he stood there with mouth open frozen in time.
     
  5. May 2, 2013 #4
    Okay... While I don't know the initial velocity, I know that it's going to hit the top of the building with a certain velocity. The velocity at 20m off the ground will be the same for both sides of the curve.
    So I can use KEi=PE
    so
    v=sqrt(2gh)
    V (at 20m) = 6.32m/s
     
  6. May 2, 2013 #5
    Took a break, but I've solved for d:

    So if I know the velocity in the Y, it turns into a trig problem. So I can find the x using tanθ=6.32/a
    So the velocity in the x axis is 3.65m/s.

    Using that, I can easily solve for D.

    3.65*4 = 14.6m

    Now I just need to solve for θ2:

    All I need to do is find the initial velocity of y, and I have enough to find θ2.

    Using Vfy = 6.32:
    6.32=vi + (-10)(2.736)
    vi=33.68
    And, again, using trig I can find the initial angle:

    tanθ=33.68/3.65
    θ2= 83.8°

    ...and I figured it out on my own! Too bad I didn't do it during the exam :(
     
  7. May 2, 2013 #6

    gneill

    User Avatar

    Staff: Mentor


    That would be true for a projectile moving vertically which just reaches 20m in height. Or for a projectile dropped (with zero initial vertical speed) from 20m upon reaching the ground. But your own diagram depicts the projectile arcing over the building edge... so it actually rises further. Can't do that if KE limits you to 20m.

    It might be worthwhile considering the trajectory in reverse (trajectories are time-reverse symmetric). Suppose the ball is launched from the roof at the given angle and reaches the ground in the given time of 4s. What's the initial y-speed? Use the appropriate kinematic expression for the y-component of the trajectory. What's the initial velocity? Carve it into x and y components.
     
  8. May 5, 2013 #7
    Great post!

    So if I were to solve it in reverse, it would be PE = PE + KE.

    All variables are known at that point, except velocity. Makes sense! It's also extremely simple...

    It also means that I have to re-evaluate this problem tomorrow.
     
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