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Best angle to shoot a projectile

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired with speed v_0 (velocity subscript zero) at an angle θ from the horizontal as shown in the figure
    5184_a_v2.jpg
    Consider your advice to an artillery officer who has the following problem. From his current position, he must shoot over a hill of height H at a target on the other side, which has the same elevation as his gun. He knows from his accurate map both the bearing and the distance R to the target and also that the hill is halfway to the target. To shoot as accurately as possible, he wants the projectile to just barely pass above the hill.

    Find the angle θ above the horizontal at which the projectile should be fired.
    Express your answer in terms of H and R.

    2. Relevant equations
    v_0x = v_0 * cosθ
    v_0y = v_0 * sinθ
    tanθ = (sinθ)/(cosθ)

    3. My attempt at a solution
    According to the description, the hill is "halfway to the target". Thus, the highest point is at R/2.
    I can then make a triangle with H as the opposite side and (R/2) as the adjacent side. Using trig, I get tanθ = opp/adj = H/(R/2) = 2H/R.

    My final answer is thus θ = arctan(2H/R).

    The answer above is incorrect and I already know what the correct answer is. What I don't understand is what's wrong with the above method? Solving it is much simpler and makes perfect sense to me. Thank you all very much.
     
  2. jcsd
  3. Oct 8, 2014 #2

    Orodruin

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    Is the projectile going to travel in a straight line up to the top of the hill and then instantly turn and do the same down to the target?
     
  4. Oct 8, 2014 #3
    How about running an example problem using given values for H,R and g (local gravitational acceleration)
    H = 10 meters
    R = 100 meters
    g = 9.81 (m/s)/s

    Find launch angle A

    Ive attached a personal crib sheet that may help, note that the horizontal velocity vector (vh) is constant (key fact)
     

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  5. Oct 9, 2014 #4
    @Orodruin I understand now. The projectile does not move in a straight line, so trigonometry can't be used. Thanks everyone for helping.
     
  6. Apr 13, 2015 #5
    what's the real answer and how do you find it?
     
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