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Physics Problem

  1. Nov 19, 2004 #1
    Click here for a picture of this problem!

    Ok as you can see the initial velocity is 6.425 m/s^2 and it is about .871m off the ground and about 3.25 m from the target. We have to find the angle at which to place the gun that shoots this ball to hit the target. This target it only like 1 inch at the largest... The gun is on top of the counter and the bullet will follow the dotted lines in the direction of the arrows, I need to know how to do this problem, and if possible the answer. Oh yeah we use 9.8 m/s^2 for gravity. Thanks in advance for any help.
  2. jcsd
  3. Nov 19, 2004 #2
    one question how is the velocity squared?
  4. Nov 19, 2004 #3
    also what are ur attempts at this?
  5. Nov 20, 2004 #4
    Velocity is not squared, that is a typo...

    Also I know it deals with projectile motion, you can split it up into two seperate problems, 1: from the gun to the level of the table, then 2: from that point to the floor. You would have to seperate the projectile motion into X and Y motion and use Trigometric functions suck as sin, cos, and tan to figure the angles.
  6. Nov 20, 2004 #5


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    That will work. The problem is that you have two unknowns for the vertical motion (Y) - you need both the angle and the time. The second part (horizontal motion) can be rearranged to solve for t, then substituted into the vertical motion equation.

    You still wind up with an annoying "sin x cos x" combination. If you're allowed to use a graphing calculator, it's pretty easy to graph that, then find your 'roots' (the points where the graph crosses the X-axis). If you have a TI there's even an option for it in the graphing menu, under 'MATH'.

    If you think about the angle that would give maximum range, you'll realize you have to have two correct answers, since both the angle and the time to hit the ground are unknown.
  7. Nov 21, 2004 #6
    Can you give me the answer so I know that I am doing to right, the two angles? Because I have to show my work so even if you give me the answer it wouldn't matter. I just really need to get this right...
  8. Nov 21, 2004 #7
    Wouldn't the time be 1.9 seconds?

    Use the equation:

    Change in Time = X displacement / Initial Velocity (Cosine Theta)
  9. Nov 21, 2004 #8
    There should be 2 diff times, shouldnt there be?
  10. Nov 22, 2004 #9
    Well what I realy need to know is the angle at which to place the gun so that it hits the target. Which means you would have to use the inverse of the trigometric functions (sin, cos, or tan). I think I might just make a program that finds the answer for me. THis isn't supposed to be an easy question. It isn't even 12th grade physics, this is a college level problem I got from Physics 1 in college. So whatever, the problem is with two unknown variables (both angles), I think you have to guess until you get them right. So I will make a program I think... unless anyone here knows how to do it easier?
    Last edited: Nov 22, 2004
  11. Nov 22, 2004 #10


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    Dearly Missed

    This is quite easy if you think a bit..
    a) Let v0 be initial speed, H vertical distance to the floor, and L the horizontal distance to the point you wish to reach.
    b) Time to reach the horizontal position of target:
    Clearly, that time [tex]\tau[/tex] fulfills:
    c) At the same instant of time [tex]\tau[/tex], the vertical position must be on the floor.
    That is:
    This gives you the equation the angle must fulfill:
    Use [tex]sec^{2}\theta=tan^{2}\theta+1[/tex] to gain a quadratic equation in [tex]\theta[/tex]
    Last edited: Nov 22, 2004
  12. Nov 24, 2004 #11
    Wow, thanks... See thats not that easy to me... lol. But with that in mind I will work through it. Thanks all of you.
  13. Nov 27, 2004 #12
    Ok, I got 13 degrees, but I think that is wrong. I am not sure how to do sec^2. You have to remember I am in america, my education isn't that great. I am useing a TI-89 to solve for theta in that equation, but like I said... I didn't even know there was a sec^2. Anyone got any other tips, thanks again in advance for any help.
  14. Nov 27, 2004 #13
    t^2 gives u

    [tex]\frac { L^2 } { (V_0 \cos \alpha)^2 } [/tex]

    so u can change 1/cos^2 to sec^2
    Last edited by a moderator: Nov 27, 2004
  15. Nov 27, 2004 #14
    I plotted the graph of the y(theta) that arildno mentioned.
    putting H = 0.871 and L = 3.25 and V = 6.425...

    I traced it using my calc... and found that when y = 0 , x = 7.05 degrees
    I hope that's right..
  16. Nov 28, 2004 #15
    Ok I plug this into my clalc:


    It gives me the answer:

    88.4233 degrees

    I don't know if that is correct or what?
  17. Nov 28, 2004 #16

    I am not sure if this is right, how can T = what that says... It would if the path was just one part, but it is two. You have to break it down into two parts don't you? Like the first part would be if you draw an imaginary line across the table, the whole motion from start to that line in the air would be one. THen the second would be free fall from that point. So t would be alot more complicated wouldn't it? If I am wrong Arildno, can you please give me an answer to this question so I can work it backwards and try to figure it out?
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