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Projectile Motion

  1. Feb 26, 2006 #1
    Here goes:

    A sniper shot is fired at over a kilometer away. Furthermore, the target is at the very edge of a building that is 40 m high. Solve for the elevation angle that will be required to score a perfect hit. The initial velocity of the bullet is 854 m/s.

    Note: the bullet is not at max height when it hits the target.

    I started out by breaking up the sniper into components. The horizontal component came out to 854cos(x) and the vertical 854sin(x).

    I solved for the time using the horizontal component; it came out to 1.756/cosx

    The time for the horizontal component equals the vertical; thus, I subbed time into d = v1(t) + 1/2(t^2).

    I got this: 40 = 854sinx(1.756/cosx) + 4.9(1.756/cosx)^2

    This is where I run into a problem. I tried to isolate for x by using various identities and nothing came through. I tried converting everything to tanx, and still no luck. I would greatly appreciate it if someone could give me a few hints. Thanks in advance,

    Nicholas
     
  2. jcsd
  3. Feb 26, 2006 #2
    I don't see where that 1.756 came from, and how did you get that equation for the time?
     
  4. Feb 26, 2006 #3

    I, too, disagree with the 1.756. As well, what is the direction of the acceleration due to gravity? You have a sign problem in your y equation. Also, is there any way to turn the sine function into an expression with only cosine in it?

    Frankly, my attack would be to solve your x equation, not for time, but for cosine and use my above suggestion to substitute it into your y equation and get an equation in the variable t. It will be a "biquadratic" equation: [tex] at^4+bt^2+c=0[/tex] for which you can solve for [tex]t^2[/tex] using the quadratic equation. You will have 4 solutions to this equation and two of them should immediately be unphysical. Then use those t values to find your angles. (Then you need to decide which angle the sniper would use.)

    -Dan
     
  5. Feb 26, 2006 #4
    the 1.756 came from this:

    Horizontal:

    v = d/t
    854cosx = 1500/t
    t = 1500/854cosx
    t = 1.756/cosx

    Since we know the time of flight for the horizontal is the same time it takes for the bullet to travel up to its maximum and then down to the target, we can substitute time into the vertical component's equation d = v1t + 0.5t^2

    The velocity is negative because it accounts for the bullet's motion; it goes up to a maximum, and then comes down to the target, which happens to be a displacement of 40 m.

    Note: I forgot to put a negative in front of the 854sinx in my first post.

    I tried your method and get stumped because I cant sub in cosx. I get something like this: 40 = -854sinx(t) + 4.9(t^2).
    Should I sub time in, then sub in cosx? Wouldn't that still leave me with a sinx? Perhaps there is some way to convert sinx to cosx, or some form of cos? Any help would be appreciated. Thanks,

    Nick
     
    Last edited: Feb 26, 2006
  6. Feb 26, 2006 #5
    Two things. You never mentioned the 1500 m in your first post! :grumpy:
    Next, the v1 in your d equation is the INITIAL velocity component in the y direction, not the final. The sniper needs to be shooting upward to get it that far away, so your v1 is +. Your number is for the component (from the first post) is correct though. What direction is the acceleration in?

    Recall that [tex]sin^2 \theta + cos^2 \theta = 1[/tex]. So solve your x equation for cosine. Then use the equation I just gave you to convert that to the sine function. (i.e. You have an equation for cosine in terms of t. Plug that into the sin^2 equation to get sine in terms of t. Then plug that into your y equation.)

    -Dan
     
    Last edited: Feb 26, 2006
  7. Feb 26, 2006 #6
    I think velocity has to be negative in this situation. The bullet travels a certain distance, reaches maximum height, and then comes down to the target. Thus, the vertical component is doing something like this:

    *|
    || <---- 40 m
    |*
    |
    |

    Haha, bear with the pathetic diagram. By making the velocity negative and the acceleration positive I am accounting for that brief split second when the bullet has a velocity of 0m/s in the air. This saves me a few calculations.

    Oh, and sorry for forgetting the 1500 m! That must have made things a bit tricky. Anyway, I still have no idea how to solve the question. Can you just show me how to solve it - step by step? I really hate doing this, but I have run out of time - my assignment is due Monday and I have to hit the algebra books. Thanks in advance,

    Nicholas
     
  8. Feb 26, 2006 #7

    Hootenanny

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    Ok. You did right by splittingit up into horiztonal and vertical components. But you dont know the actual horizontal displacement, it is simply 'over a km'. However, if you assume that the sniper is at ground level then you do know the vertical displacement. From this you can calculate the flight time.
     
  9. Feb 26, 2006 #8
    hey, im trying to interpret his words into an image, and ive come to this.

    [​IMG]

    I think the acceleration is negative, is this the right picture of what he is asking.. and your calculating the time the bullet is in the air from when it is launch and until it hits the red dot?
     
  10. Feb 26, 2006 #9
    I'm not sure that actually helps, but as long as you keep track of what you are doing it's fine. Also, note then in your coordinate system the final vertical displacement will be negative.

    With regards to MrRottenTreats, we posted at the same time! You CAN define the positive y direction to be downward as long as you keep the signs consistent. I note on your picture you have the vector a as negative, which is not correct. The acceleration vector points downward, but vectors carry no sign themselves. Only the components of a vector carry signs. Thanks for the diagram, though. I've really got to get my scanner working!

    -Dan
     
    Last edited: Feb 26, 2006
  11. Feb 26, 2006 #10

    Hootenanny

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    Your image above is incorrect, because it states in the question that the bullet has not reached its maximum height when it strikes the target.
     
  12. Feb 26, 2006 #11
    No the question says the bullet is not at its maximum height when it hits. But where is the 1500 meters coming from where does it say that in the original problem?
     
  13. Feb 26, 2006 #12

    Hootenanny

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    Oops mistake. Yes I can't find the 1.5km in the problem. That is why I suggested using the vertical displacement.
     
  14. Feb 26, 2006 #13
    okay.. hmm so my diagram is just a rough sketch, asking him if it loooks like this. the 1.5 k, the 40 m as the height and we need to find the angle and time?

    but i dont think the bullet is at its max height when it hits the target...

    can we asssume that the bullet is at its max height when it hits the red dot?
     
  15. Feb 26, 2006 #14

    Hootenanny

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    Where did the 1.5km come from? Apart from that it looks good.
     
  16. Feb 26, 2006 #15
    It's good that everyone is grasping the concept. I forgot to put that the distance from the sniper to the target is 1500 m, which is not the entire distance travelled, but merely the distance from the sniper to the target.

    The distance from the sniper to the target is 1500 m

    Ok, now it should be a bit easier. Anyone have an idea as to how I can solve for the angle? Maybe my givens are flawed?

    Horizontal:

    v = 854cosx
    d = 1500m
    t = d/v

    Vertical

    a = 9.8 m/s^2
    d = 40m -I figure the bullet will hit maximum, then fall and hit the target. Thus, the displacement of the bullet will be 40m).
    v1 = -854sinx
    t = t(horizontal) - the time should be equal.

    Does this make sense? I made the velocity negative to account for the bullets vertical motion; it will go up, hit max height, then drop down to the target. Its displacement should therefore be 40 m.
     
  17. Feb 26, 2006 #16

    Hootenanny

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    Your logic is correct. Thank's for adding the extra information. Except your vertical acceleration should be negative as your displacement is positive. Its initial velocity (the one which you are calculating) is positive, as it is going up.
     
  18. Feb 26, 2006 #17
    i think the 1500 is the distance from where the target, 40 m high is located, which we must hit.
     
  19. Feb 26, 2006 #18

    Do me two favors, Hadjiev. First use different variables when you are doing the x and y directions. It's too easy to get things confused. Also, you STILL are not conforming to your coordinate system.
    The problem states that the final vertical displacement component is 40 m. Now, look at MrRottenTreats' diagram. 40 m means that the target is 40 m above the sniper. In your coordinate system you have + downward. Okay. But if + is downward and you have something 40 m ABOVE you, which side of the y-axis is that? I'll state it explicitly: In your coordinate system d = -40 m!

    -Dan
     
  20. Feb 26, 2006 #19
    hey hold, are we solving for the time or the angle or both?
     
  21. Feb 26, 2006 #20

    Hootenanny

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    Just the angle mate.
     
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