Maximum range of a projectile when launched at 45°(Physics by example, Rees, pr. 14)

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    In this solution the t in the y-coordinate equation is substituted using the x-coordinate equation and ultimately leads to the answer.

    My questions:
    1. Why don't I get the same answer when I substitute the v or v and t instead?
    2. How am I supposed to know to substitute t in this example and not v?
     

    Attached Files:

  2. jcsd
  3. Delphi51

    Delphi51 3,410
    Homework Helper

    Re: Maximum range of a projectile when launched at 45°(Physics by example, Rees, pr.

    I don't see how you can get it by keeping the t. You don't know t, so how can you tell the maximum value of x that way? I get
    x = gt²/(2*tanΘ)
    and can't tell what combination of t and Θ provide the maximum x.
    It would be good to see your calc.

    There is no need to eliminate the v; it is a constant. But you must eliminate the variable t.
     
  4. Re: Maximum range of a projectile when launched at 45°(Physics by example, Rees, pr.

    Well, the problem is quite easy to approach.
    Range is given by u^2 sin(2a) / g, where a is the projection angle.
    Since -1<sin a<1, the max. value for a sine function = 1. This occurs when the angle is 90 degrees or .5pi radians.
    So, for a fixed u:
    2a = 90
    a = 45 degrees.
     
  5. Re: Maximum range of a projectile when launched at 45°(Physics by example, Rees, pr.

    Ok, now I understand. Thnx!
     
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