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2 Projectile Motion Problems

  1. Jun 11, 2012 #1
    1. A soccer player kicks a ball at an angle of 50° above the horizontal. The player is 10.0 m from the base of a rectangular building that is 5.0 m high (with a flat roof). The roof is 10.0 m wide. The ball is kicked with an initial speed of 15.0 m/s.
    a.Show that the ball will clear the front wall of the building.
    b.Does the maximum height of the ball occur while it is over the roof, before it passes over the roof, or after it passes over the roof?
    c.Determine whether the ball lands on the roof or beyond the building.
    d.Wherever the ball lands, determine its location relative to the back wall of the building.
    e.What is the ball’s velocity in unit vector notation, just before it lands?


    WHAT I DID TO GET THE ANSWERS
    i got these answers. I can't really show my work since I don't have a scanner or smartphone to take a pic but I did do this and tried it myself.
    i got x component of initial velocity= 9.64 m/s, y component of initial velocity= 11.49 m/s, t= 1.17 s for it to reach max height using the kinematic equations for projectile motion.
    I used these known variables to find the answers to a and b. then I set the height to 5 and found t for that in order to find the range covered at that point for c. Then I subtracted that figure from the total distance from the soccer player to the end of the building to find d. As for e, i don't know how to do it because i don't know how to find y.
    a. max height ---> y=6.73 m so it goes over the wall. @ y=5 m, x is about 5.4 meter
    b. x=11.3 m so it is while it is over the roof
    c. x=17.045 m so it lands on the roof
    d. x=2.95 meters before the back wall of building. i did 20- 17.045
    e. i don't know how to do this
    idk how to do this since x is constant and idk how to figure out the velocity of y

    i just need help with e.
    I figured out the other parts. thanks :)

    2.JUST FIGURED THIS ONE OUT. I don't need help with it anymore. In this experiment: h ttp://www.youtube.com/watch?v=cxvsHNRXLjw, prove that the pellet does hit the doll (provided it is shot fast enough).
     
    Last edited: Jun 11, 2012
  2. jcsd
  3. Jun 11, 2012 #2

    Simon Bridge

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    What is the relationship between the velocity of the ball at the end of it's flight and that at the start?

    Note: just to clear up your notation...
    "velocity of y" is not correct - it should be the y component of the ball's velocity - same with x. In fact, the figures you show are the components of the initial velocity - after all, the actual y-component of velocity changes with time. Better to call them horizontal and vertical components in this case - helps you reason as well as communicate.

    If +y is "up" then that is the j direction.
    Your equation renders: [itex]\Delta y = v_{y0}t - \frac{1}{2}t^{1/2}[/itex] ... which looks odd to me: why the square root (power half) in there and what about gravity?
    I have found the quickest check is to draw a velocity-time graph for each component.
     
    Last edited: Jun 11, 2012
  4. Jun 11, 2012 #3
    well the initial velocity is 15 m/s and when the ball lands on the roof, that velocity decreases to 0. the acceleration is constant though.

    oh thanks for the note on the notation. i'll clear that up
     
    Last edited: Jun 11, 2012
  5. Jun 11, 2012 #4
    e.What is the ball’s velocity in unit vector notation, just before it lands?

    It is the unit vectors in the coordinates axes.

    Fx=Fxi
    Fy=Fyj
    Fz=Fzk
     
  6. Jun 11, 2012 #5

    Simon Bridge

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    I'm sorry - the ball lands on the roof ... so what you need is the speed horizontal and vertical when it does this. Can you work out an equation of speed vs time for each component?

    The question asks - "just before" it lands because when it lands it bounces and all bets are off.
    Note: the initial speed is 15m/s - what is the initial velocity? Do you know how to write it in unit vector notation?
     
  7. Jun 11, 2012 #6
    well the equation for the x component is delta x= x component of the initial velocity * t

    and i think, the equation for the y component is --> y component of the final velocity= y component of the initial velocity *t - gt
     
  8. Jun 11, 2012 #7
    well, unit vector notation is xi +yj+ zk right?
     
  9. Jun 11, 2012 #8
    so would the answer be 9.64i -5.83769j?
     
  10. Jun 11, 2012 #9
    i got x component of initial velocity= 9.64 m/s, y component of initial velocity= 11.49 m/s, t= 1.17 s for it to reach max height using the kinematic equations for projectile motion.
    ................................

    For question (e) you have indirectly already got the answer.
    Wonder how you get t=1.17sec.
     
  11. Jun 11, 2012 #10

    i just realized what i did. i was solving for t when the ball reached maximum height when i was supposed to set the height to 5. thanks for pointing that out
     
  12. Jun 11, 2012 #11

    Simon Bridge

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    All right you seem to have the basic idea - lets tidy up:

    let u = 15m/s is the initial speed of the ball.
    The initial velocity is [itex]\vec{v}(t=0) = \vec{i}u\cos\theta + \vec{j}u\sin\theta[/itex]
    (quote me to see how I wrote that - you should learn how as well)

    [itex]v_{x}(t) = u\cos\theta[/itex] ... ... (1)
    [itex] x(t) = ut\cos\theta[/itex] ... ... (2)
    [itex] v_{y}(t) = u\sin\theta - gt[/itex] ... (3)
    [itex] y(t) = ?[/itex] ... ... (4)
    ... finish equation 4: it has to be the time-integral of the velocity, remembering that y(0)=0. You can also work it out from kinematic equations or from the v-t diagram.

    Questions (a) and (c) ask about the height of the ball at particular positions - to solve them you need y(x) which you can get from equations (4) and (2). You already know what shape this should be right?

    Question (b) wants the x position of the maxima of y(x).

    You get the idea?
    Question (e) needs you to know the speeds when y is equal to the height of the building ... which would need [itex]v_x(y)[/itex] and [itex]v_y(y)[/itex] ... or just use equation 4 to solve for the time, then plug the time into (1) and (3). [which is actually what you've done]
     
    Last edited: Jun 11, 2012
  13. Jun 11, 2012 #12
    awesome thanks! I figured it out after aziz pointed out my error for part a.
    thank you for your help! its just a quadratic equation. lol i can't believe it took me this long to figure this out.
     
  14. Jun 11, 2012 #13

    Simon Bridge

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    Well done - helps if you list everything you know and keep your terminology clear.
     
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