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If we neglect air resistance, then the range of a ball

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  1. Oct 12, 2015 #1
    1. The problem statement, all variables and given/known data
    If we neglect air resistance, then the range of a ball (or any projectile) shot at an angle θ with respect to the x axis and with an initial velocity v0, is given by
    R(θ)=v_0^2/g sin(2θ) for 0 ≤ θ ≤ π/2

    where g is the acceleration due to gravity (9.8 meters per second per second).
    For what value of θ is the maximum range attained? (Note that the answer is numerical, not symbolic.)
    θ =

    3. The attempt at a solution
    If I understand correctly. To find the max and min you simply take the derivative of that function and set it to zero. In this case I assume we have only a max since a ball is being tossed into the air.

    I get the functions derivative:
    R'(θ) = V_0^2/g *2*cos(2θ)
    Then isolate the function
    cos(2θ) = 0, well that is only true for pi/2, since it's domain here is [0,pi/2]
    therefore isolating θ by itself by dividing through.
    2θ=pi/2

    θ=pi/4

    Is this process and reasoning correctly thought out?

    Thank you
    Os
     
    Last edited by a moderator: Oct 12, 2015
  2. jcsd
  3. Oct 12, 2015 #2

    Mark44

    Staff: Mentor

    Looks good to me. For maximum range, the angle should be ##\pi/4##.
     
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