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Projectile motion problems

  1. Jun 24, 2012 #1
    Can someone tell me how to do this problem? I know 16. is A, 17 is E? 3rd pick is of what i have done not even sure if right.
     

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    Last edited: Jun 25, 2012
  2. jcsd
  3. Jun 25, 2012 #2
    horizontal velocity is constant = [itex]v_{x}=v_0 cos(53°)[/itex]
    horizontal distance covered at any time [itex]t[/itex] is [itex]S_x = v_{x}× t = v_0 cos(53°)×t[/itex]
    which gives time to cover [itex]S_x( = 25) = \frac{25}{v_0 cos(53°)} =t [/itex]
    initial vertical velocity [itex]v_y = v_0 sin(53°)[/itex]
    vertical distance traveled [itex] 12 = v_0 sin(53°) ×t - \frac{1}{2} gt^2[/itex]
    substituting the value of [itex]t[/itex] and simplifying we get

    [itex] v_0 = \sqrt {\frac{g×25^2}{2(25 tan(53°)-12) cos^2(53°)}}≈20 m/s[/itex] [taking g = 9.8m/s^2]

    so [itex]v_x = 20 × cos(53°) ≈ 12 m/s [/itex]
    etc.....
    you can do the rest with all the formulas
     
  4. Jun 25, 2012 #3
    You found out ,

    T= 25/vox

    vox = 14

    Are you asking 18 , 19 , ....22.

    For 18 :

    You are to find voy

    Putting vox = 14 in T= 25/vox , find numerical value of T. Then putting formula of time of flight in a projectile you can solve for vo. Then you can find voy.

    Other approach is that
    vocosθ= 14
    vosinθ=z
    On dividing ,
    cotθ = 14/z
    solve for z....
     
    Last edited: Jun 25, 2012
  5. Jun 25, 2012 #4

    Simon Bridge

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    spot the 3-4-5 triangle ;)

    given initial and final displacements:
    yi=xi=0; yf=12m; xf=25m

    from the 3-4-5 triangle:
    3v0y=4v0x ...1

    time of flight
    T = 25/v0x ...2

    12 = v0yT - gT2/2 ...3

    three equations, three unknowns.

    One more equation comes from the slope of the vy vs t graph.
     
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