1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projectile Motion Problem, need Help

  1. Nov 12, 2011 #1
    Here is the problem :
    A Sniper is lying in a prone position such that the barrel of his rifle is parallel to and 30cm above the ground. The muzzle velocity of his rifle is 800m/s

    A target appears at a distance of 1250m with the centre of the target 1.2m above ground level. If the sniper were firing from the prone position, calculate the angle to which the barrel of the rifle should be raised to hit the centre of the target

    Also there is assumed no air resistance.

    The Equations for projectiles are:
    Vx = Ux
    v = u + at
    Vy^2 = Uy^2 + 2ay * Δy (ay is -9.8, don't know how to type subscript)
    Δx = Ux * t
    Δy = Uy * t + a/2 * t^2
    Could someone please explain how to solve this problem

    I'm in highschool and this question is part of my HW btw
     
    Last edited: Nov 12, 2011
  2. jcsd
  3. Nov 12, 2011 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Do you understand sine and cosine?

    The bullet emerges from the barrel and follows a curved path, just as does a thrown tennis ball, except the bullet does it faster. So, horizontally, the bullet's speed remains constant, because there is no horizontal force that can change it. Vertically, gravity is constantly dragging the bullet downwards, so you have to allow for this acceleration in the vertical component of speed so that by the time it reaches the target the bullet has fallen groundwards to the extent that it now intercepts the target.
     
  4. Nov 13, 2011 #3
    Yes i have a sufficient knowledge of trig

    Ultimately the answer needed is θ,

    and i've been trying to get it through solving time.

    Through the formula Δx = Ux * t
    i have solved t to being 1250/800cosθ

    When i substitute this into Δy = Uy * t + a/2 * t^2 where Δy is 0.9m (1.2 - 0.3), a is -9.8 & Uy is 800sinθ
    it gives me some crazy unsolvable equation

    [PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP56819i2ceg3cd7aic460000264e3bge3g75cdf0?MSPStoreType=image/gif&s=24&w=373&h=38 [Broken]

    Wolframalpha tells me the answer is
    [PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP50819i2cfib8ae7agfb000042ifb2da9b30b08b?MSPStoreType=image/gif&s=46&w=500&h=22 [Broken] and i have no idea what that is

    Surely the angle is a real number and this is wrong
    is my t wrong, or is it something else?
     
    Last edited by a moderator: May 5, 2017
  5. Nov 14, 2011 #4
    Make use of the fact that V2 = Ux2 + Uy2
     
  6. Nov 14, 2011 #5

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Split the problem into two steps:
    You can work out the horizontal component of velocity (with unknown angle θ). This will give you an expression for the time of flight over the total distance.
    You then look at the vertical motion (choose the right equation), substitute the time value that you just worked out. This will yield an equation which will give you a 'sensible' value for the θ you need.
    It WILL give the right answer if you are careful about the maths manipulation.
     
  7. Nov 14, 2011 #6
    Thanks but Using this. i get 8002 = 8002sin2θ + 8002cos2θ
    = 8002(sin2θ + cos2θ)
    which is 8002 = 8002(1) because sin2θ + cos2θ = 1 for any theta
    making this a useless equation


    Ok so, using trig, i get Ux = 800cosθ
    substituting this into Δx = Ux * t gives
    1250 = 800cosθ * t
    t = 1250/800cosθ

    Now substituting t into Δy = Uy * t + a/2 * t2
    i get 0.9 = 800sinθ ( 1250/800cosθ ) - 9.8/2 * (1250/800cosθ)2
    0.9 = 1250sinθ/cosθ - 4.9 ( 12502 / 8002cos2θ )

    From here i find it impossible to solve for theta, could someone explain where i went wrong.

    Thanks
     
    Last edited: Nov 14, 2011
  8. Nov 14, 2011 #7

    Delphi51

    User Avatar
    Homework Helper

    I'm the old high school teacher here. It is surprising what can be solved with high school methods! We write the horizontal d = vt: 1250 = 800*cos(θ)*t [1]
    And the vertical d = Vi*t+.5*a*t²: 0.9 = 800*sin(θ)*t - 4.905*t² [2]
    Note that we have two equations and unknowns t and θ. Solve [1] for t and
    substitute into [2] to get 0.9 = 800*1.5625*tan(θ) - 11.975/cos²(θ).
    I think that is the same equation you have!
    I can't solve that with high school trig, but the kids can get it on their graphing calculators instantly by drawing the graph of the right side and seeing when it is equal to 0.9. I put the right side into a spreadsheet cell, then tried various values of θ. It seems that θ = 0.5902 degrees solves it.
     
  9. Nov 14, 2011 #8
    How did you get 1.5625 and 11.975?
     
  10. Nov 15, 2011 #9

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I have not checked your long equation but, if you want to solve it, you should be able to reduce it to an equation in Cos θ, which should then be soluble. (Use the identity for sin to cos that you wrote earlier )
     
  11. Nov 15, 2011 #10

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    The equation for the trajectory is a parabola which is only, basically, a quadratic. It's just a matter of finding where what you know fit into a quadratic.
     
  12. Nov 15, 2011 #11

    Delphi51

    User Avatar
    Homework Helper

    t = 1250/800cosθ = 1.5625/cosθ

    0.9 = 800*sin(θ)*t - 4.905*t²
    0.9 = 800*sin(θ)*1.5625/cosθ - 4.905*(1.5625/cosθ)²
    0.9 = 1250*tan(θ) - 11.975/cos²θ

    I see your idea, Sophie. And it would be more satisfying to get the answer without resorting to guess & test. But it looks difficult.
     
  13. Nov 15, 2011 #12

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Did you know that secsquared is 1+tansquared? If you don't know sec theta then look it up and you will find it leads to a quadratic in tan theta.
     
  14. Nov 15, 2011 #13

    Delphi51

    User Avatar
    Homework Helper

    Great insight! I used to know that identity but it is no longer on instant recall.
     
  15. Nov 15, 2011 #14

    NascentOxygen

    User Avatar

    Staff: Mentor

    Time of flight is around a minute and a half, so it's a slow old bullet! That would surely be about the speed of one of Sir Francis Drake's cannon balls? :smile:
     
  16. Nov 16, 2011 #15

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Are you sure that you subbed correctly? Where did that square root come from? ;-)
     
  17. Nov 16, 2011 #16
    According to wikipedia,
    Using the pythagorean identity for cosθ in terms of tanθ equals
    c52bb4d9c170f748c9f8f204ee3ae1d3.png

    Subbing this into 0.9 = 1250*tan(θ) - 11.975/cos²θ
    gives
    0.9 = 1250tanθ - 11.975/ (±1/√(1 + tan2θ))2
    0.9 = 1250tanθ ± 11.975 * (1 + tan2θ)2

    oops, looks like i wrote it wrong...,
    WOW! i this looks solvable!!!!

    expanding gives 0.9 = 1250tanθ -1(11.975 + 11.975tan2θ) //This looks an awful lot like a quadratic

    rearranging gives

    11.975tan2θ - 1250tanθ + 12.875 = 0

    Solving this using the quadratic formula yields

    tanθ = [1250 ± √(12502 - 4(11.975 * 12.875))] / [2(11.975)]

    Using arctan gives

    89.45106906 & 0.5901838982

    This is the correct answer!!!!!!!!!!
    Finally! Thanks man
     
  18. Nov 16, 2011 #17

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Well done. Big relief!!!
    btw, I would have just subbed the 1/cossquared with 1 - tansquared all in one go.

    We had to learn a lot of these identities at School in the Middle Ages - good old Mr Worthington, the legend. I have forgotten them all, nearly, but I always look at the list because there's often a useful one in there somewhere.
     
  19. Nov 16, 2011 #18
    Here's what I meant:

    Using ΔX = 1250, ΔY = 0.9, U = 800, a = 9.8, you have
    [tex]U_x \cdot t = 1250[/tex]
    and
    [tex]U_y \cdot t - \frac{1}{2} 9.8 \cdot t^2 = 0.9[/tex]

    Solving each for U gives
    [tex]U_x = \frac{1250}{t}[/tex]
    and
    [tex]U_y = \frac{4.9 t^2 + 0.9}{t} = 4.9t +\frac{0.9}{t}[/tex]

    Squaring both gives
    [tex]{U_x}^2 = \frac{1250^2}{t^2}[/tex]
    and
    [tex]{U_y}^2 = 4.9^2 t^2 + \frac{0.9^2}{t^2} + (2) (4.9)(0.9)=24.01t^2+\frac{0.81}{t^2} +8.82[/tex]

    But, since we know that U2 = Ux2 + Uy2, we have:

    [tex]800^2 - {U_x}^2 = 24.01t^2 + \frac{0.81}{t^2} + 8.82[/tex]
    which gives us
    [tex]800^2 - \frac{1250^2}{t^2}= 24.01t^2 + \frac{0.81}{t^2} + 8.82[/tex]

    Multiply by t2 throughout and combine terms to get
    [tex]24.01t^4 + (8.82 - 800^2)t^2 + (0.81 + 1250^2) = 0[/tex]
    which is a quadratic in t2

    Simply solve for t, from which you can get Ux, Uy, and therefore θ

    No trig identities needed!
     
  20. Nov 16, 2011 #19

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    OK but then you need to find the angle needed. (That's the question)
     
  21. Nov 16, 2011 #20
    Yeah, like I said, "Simply solve for t, from which you can get Ux, Uy, and therefore θ"

    This can be done by using
    [tex]U_x \cdot t = 1250[/tex]
    Remember that
    [tex]U_x = U \cdot cos\theta[/tex]

    So,
    Plug in the values of t and U into the the equation and solve:
    [tex]800 t \cdot cos \theta = 1250[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Projectile Motion Problem, need Help
Loading...