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Projectile motion football problem

  1. Dec 9, 2006 #1
    Hey guys, im studying for my physics exam and i have not been able to get this question for quite a while now!

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

    A football is kicked at an angle of 50 degrees to the horizontal and travels a horizontal distance of 20.0m before hitting the ground. Find:
    a) the initial speed of the ball
    b) the time it is in the air
    c) max height it reaches

    3. The attempt at a solution

    now i i know wi need to split the 50 degrees and the speed into its x and y compenents, and i will only be dealing the with Vcos50 portion of it, but from there im stumped.. i am awqare that there will be subbing of questions but i just cant get it
    please help me out guys!
     
  2. jcsd
  3. Dec 9, 2006 #2

    cristo

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    you're correct in thinking you need to split into x and y compents. Try considering the first half of the motion, ie from when the particle is launched, to when it reaches its maximum height. Set up the "suvat" equations in the x direction. We know that the x component of the initial velocity is vsin50. Do you know what the final velocity is when the particle reaches its maximum height? Also, if you recognise what the aceleration is, then you should be able to calculate v and the maximum height. Post back when you have solutions to this, or working if you have problems
     
  4. Dec 9, 2006 #3
    ok.. well i know that the acceleration is -9.81. now, ur telling me to approach it by thinking of it halfway, ok so the vf at the peak height is zero.
    now..i just odnt know what the "suvat" equation is.. i prolly do but just not by that name
     
  5. Dec 9, 2006 #4
    and isnt the x component vcos50?

    and so far i have 20= Vcos50t + -4.9 t^2
     
  6. Dec 9, 2006 #5
    Do you know the forumlas for uniform acceleration?
    We dont consider horizontal component of velocity because acceleration due to gravity has no component in the x direction. It remains constant throughout the motion. However its vertical velocity is affected due to acceleration due to gravity in the y direction

    (a) The initial speed of the object is its vertical component in the y direction

    (b) If you know that displacement is zero can you find t? if you have its initial vertical component of velocity. Do you take the acceleration due to gravity positive or negative?
    hint: considering vertical motion

    (c) What is the velocity of the ball when it has reached maximum height, if so
    can you find its maximum height if you know its initial vertical component of velocity.
     
    Last edited: Dec 9, 2006
  7. Dec 9, 2006 #6

    cristo

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    yes, sorry, in my previous post, i meant the y direction!!

    so, set up equations in the vertical (y) direction. We know that
    s=s_max
    u=Vsin50
    v=0
    a=-9.81
    t=time taken to reach max height.

    the suvat eqns are:
    v=u+at s=vt- (at^2)/2 s=t(u+v)/2 v^2=u^2+2as.

    From these you should be able to formalate 3 expressions involving s, t and V, which you can solve to obtain V and s_max
     
  8. Dec 9, 2006 #7
    im still kinda lost.. im sorry :redface: this takes a while for me to understnad

    the thing is that i dont my s max? all they give me is the 20m horizontally
     
  9. Dec 9, 2006 #8

    cristo

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    s_max is the max height of the ball (the answer to (c)). Plug into the above equations u=Vsin50, s=s_max. you should use
    v=u+at (1)
    v^2=u^2+2as (2)
    s=ut-(at^2)/2. (3)

    You correctly identified a=-9.81, v=0. so (1) becomes Vsin50=9.81t
    (2) becomes 0=(Vsin50)^2-19.62s_max
    (3) s_max=vsin50 t +4.905t^2

    use the new form of (1) to eliminate t in (3). then you will have two equations which you can solve for s_max and V
     
  10. Dec 9, 2006 #9
    Whats the formula for Range?
    Can you derive another equation for range if you replace t
    by the time the ball was in the air (considering vertical motion)

    And please cristo, try not to solve the question partially. Its against forum rules!
     
    Last edited: Dec 9, 2006
  11. Dec 9, 2006 #10
    after all this plugging in ans stuff i end up with some equations,

    v^2sin50- 2v^2sin50- 9.81v^2sin50=0 :grumpy:

    and what i did was solved equation one for T, subbed the t equation into the equation 3, whihc gave me an equation for whihc i then put into equation @ by subbing in the x
     
  12. Dec 9, 2006 #11

    cristo

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    ok, do you know how to solve this? (remember sin50 is just a number- use your calculator!!)
     
  13. Dec 9, 2006 #12
    yes ok well I STILL DONT GET AN ANSWER, BECAUSE I HAVE THE V SQUARES IN FRONT OF ALL THE SIN50'S... AND SORRY ABOUT THE CAPS MY PC IS MESSED UP
    AND TRY IT URSELF, BECAUSE LIKE I CANT GET IT TO EQUAL 14.1, WHICH IS THE ANSWER, MAYBE IM DOING A MATH ERROR
     
  14. Dec 9, 2006 #13
    Do you know that
    (vsintheta)^2=V^2sin^2theta

    And use the method that I told you.
    Derive another equation for range, by replacing t
    by the time the ball was in the air (considering vertical motion)

    R=vcostheta x T (replace this t)

    I think you need to go to the basics first, what will be the balls vertical displacement as it reaches maximum height and returns back to the ground?
    Can you find T, by using the equation s=ut+1/2at^2
     
    Last edited: Dec 9, 2006
  15. Dec 9, 2006 #14
    no but how wud that help me
     
  16. Dec 9, 2006 #15
    the balls displacement would be zero and i wud have to sub in the t=vsin50/9.81 into that equation?
     
  17. Dec 9, 2006 #16
    Yes your right, but remember thats time to reach max height. Substituting into that equation would be wrong because its vertical displacement in this time is not zero to reach maximum height.

    you have to find t though, not substitite it. Its initial vertical velocity is VSin50. Now find t in terms of vsintheta and g.

    You dont have the value of t, so by using this method, you are eliminating t from the equation. This saves a lot of hard word in your exam!.
    0=(usintheta)(t)-1/2gt^2
    t=2usintheta/g

    r=ucostheta x t
    r=ucostheta x (2usintheta/g)
    r=u^2sin^2theta/g

    Can you find u in this equation?
     
    Last edited: Dec 9, 2006
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