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Homework Help: Can someone elp me with a couple of homework questions im having trouble with

  1. Apr 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A hunter aims directly at a target (on the same level) 54.08m away. If the bullet leaves the gun at a speed of 143.91 m/s then at what angle should the gun be aimed in order to hit the target? (answer in radians)

    2. Relevant equations

    3. The attempt at a solution
    Vx = Vcos theta, Vx/V cos-1 = theta
    im pretty sure this is not the right way to do it but i just dont know how to start or what equations to use
  2. jcsd
  3. Apr 23, 2008 #2
    Involves projectile motion.
  4. Apr 24, 2008 #3
    so I used this equation to y=v^2y0-v2y/2g. I rearanged it using the values i had to work out vy0, which i got 32.56, i used the distance as y and v2y as 0. Then i put that into this equation vy0=v0sin theta, to work out theta. I rearranged it to vy0/v0 sin-1 = theta, i used my value for vy0 and for v0 i used the velocity and got the answer 0.23. I then changed that to radians and got the answer 3.98E-3. But that is the wrong answer....What have I done wrong??
  5. Apr 24, 2008 #4


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    Homework Helper

    Which equation is this? It does not appear to have the right dimensions. (The term on the left has units of length; the first term on the right has dimensions length^3/time^2.)

    I don't think you have enough information to find [itex]v_{0y}[/itex] yet. Once we find the angle [itex]\theta[/itex] that this problem is asking for, then you'll have [itex]v_{0y}=v_0 \sin\theta[/itex] (if [itex]\theta[/itex]) is the angle from the horizontal).

    You have to keep track of motion separately in the horizontal (x) and vertical (y) direction.

    In this problem, what is the displacement in the horizontal direction [itex]\Delta x[/itex]? in the vertical direction [itex]\Delta y[/itex]?

    Once you have those, you may have a formula that your book derived for this special case. If so, you can calculate the angle with just that one formula.

    (If you don't have this special case formula, then you can solve the problem directly using the kinematic equations for the x and y directions.)
  6. Apr 28, 2008 #5
    i found this equation. R= v^2 sin 2 theta/g. If i put my values in there i get 2.23 x 10 ^4, the right answer is 1.28 x 10^-2. Im pretty sure this equation is the correct one, can you please give me hints on what ive done wrong?
  7. Apr 28, 2008 #6


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    Hi inflames829,

    Yes, that is the correct question. (It is derived by setting [itex]\Delta y=0[/itex] for projectile motion with no air resistance, and so applies when that is true [itex]y_f=y_i[/itex].)

    However, the answer you got is huge! The problem is asking for an angle to aim a gun, so it's looking for an answer whose magnitude is less than [itex]\pi/2[/itex] or about 1.6 or so (in radians). Did you solve your equation for theta? If so, please post what you actually put into the calculator.
  8. Apr 28, 2008 #7
    ive got the right answer now. I realise what i did wrong. Thanks for the help though
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