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Projectile Motion - Seems easy.

  1. Oct 24, 2006 #1
    Projectile Motion -- Seems easy.

    A projectile is shot from the edge of a cliff 245 m above ground level with an initial speed of vo=135 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-44.


    Figure 3-44

    (a) Determine the time taken by the projectile to hit point P at ground level.

    (b) Determine the range X of the projectile as measured from the base of the cliff.

    (c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Assume the positive directions are upward and to the right.)
    m/s (horizontal)
    m/s (vertical)

    (d) What is the magnitude of the velocity?

    (e) What is the angle made by the velocity vector with the horizontal?
    ° below the horizontal

    I tried doing \Delta X = Vox*t for A but I know that won't work because 245 is \Delta Y I can't solve for time without either \Delta X or Voy which is also what I don't have.. any hints?
    Last edited: Oct 24, 2006
  2. jcsd
  3. Oct 24, 2006 #2
    Sorry I don't know how to enter equations here, so I'll do the best I can.

    Lets start with;
    y=((v0) sin(θ))t - 1/2gt^2
    defines the y location of the projectile. Set this to ground level and we have;
    -245 = (135 sin(37))t - 1/2(9.8)t^2
    0 = 245 + 81.245*t - 4.9*t^2

    I don't have a calculator handy, but solve that for t and you know the time it took for the projectile to hit the ground.

    x=((v0) cos(θ))t give the x location if you enter the time value from above.

    At impact the horizontal component is the same as when it left (assuming no air resistance);
    v(x) = (v0)cos(θ)

    The y component is;
    v(y) = (v0)sin(θ) - gt

    velocity at impact is;
    v = sqr( (v(x))^2 + (v(y))^2 )

    lastly, the angle at impact is;
    θ = arctan( v_y/v_x) )

    I'm a student also, so please check my work. Also, how can formula's be better input so they look nicer?

    Last edited: Oct 25, 2006
  4. Oct 24, 2006 #3
    -245 = 81.245*t - 4.9t^2

    I got .784.. I got that by

    -245/8.245 then dividing the quotient of that by -4.9.. then square rooting that.

    Which is incorrect.

    I see this is your first post, thank you for your help.. if I did something wrong let me know.
    Last edited: Oct 24, 2006
  5. Oct 24, 2006 #4
    To make formulas look better you type "[..tex..]" formula stuff "[../tex..]" without the quotes or the periods.
  6. Oct 24, 2006 #5


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    Homework Helper
    Gold Member

    Bernie's method and solution is correct. You messed up on solving for t. This is a quadratic equation. Solve for t using the quadratic formula.
  7. Oct 24, 2006 #6
    Ah lame! Alright let me try.
  8. Oct 24, 2006 #7
    Okay I got 19.187 for A and it worked. I then did the formulas to get B and C

    B was 2.068km
    C was 107.815..

    I'm stuck on D though

    for v(y) = (v0)sin(θ) - gt

    I don't know what to do with that

    I tried doing 135sin(37) - 9.81(19.187)

    and that does not work.
  9. Oct 25, 2006 #8
    Hmm... in letter C, you are asked to get the x & y components... so once you got the components... get the magnitude by V=sqrt(Vx^2+Vy^2)

    to get the angle, just get the inverse tangent of the quotient of the y-component and x-component

    because tanθ=Vy/Vx
  10. Oct 25, 2006 #9
    Yeah for C2, the second C.. I can't get the component with the formula he gave me.. it doesn't add up right.
  11. Oct 25, 2006 #10
    C2? The x component, you mean? It stays the same--the vertical component (v_x) is constant at all point, so v_x = v_fx = v_ix. Just plug that in the formulas Bernie gave you and you should get the right answer. And I believe you incorrectly calculated (a) if you got 19.817--you should get 10.63 seconds.
  12. Oct 25, 2006 #11
    No, my webassign automatically checks the answer and it said it's correct at 19.87. X is horizontal, Y is vertical.
  13. Oct 25, 2006 #12
    Okay I got c2 using bernie's formula. Thanks. :)

    Trying the rest.
  14. Oct 25, 2006 #13
    Very much sorry, Kildars. I neglected the (1/2) in the equation. :blushes: Good luck with the problem!
  15. Oct 25, 2006 #14
    Got them all, thanks for the help guys.
  16. Oct 25, 2006 #15
    No problem!
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