physics -- projectile motion

posted by ashley --- HELP PLEASEE!!!

why does a projectile of 45 degrees have the greatest range?

  1. jai

    recall that the formula for range is:
    R = (vo)^2 * sin (2*theta) / (2g)
    vo = initial velocity
    theta = angle of release
    g = acceleration due to gravity (9.8 m/s^2)

    now, if we're given the initial velocity, the value of range will depend on theta,, observing this, if we want to have a max Range, we have to find the max value of sin (2*theta),, but the max value of sine is always 1, therefore, equating this to 1:
    sin (2*theta) = 1
    2*theta = arcsin 1
    2*theta = 90 degrees
    theta = 45 degrees

    hope this helps~ :)

  2. jai

    oops, sorry the formula for range must be
    R = (vo)^2 * sin (2*theta) / g

    i just remembered that the 2g is for the max height:
    h,max = (vo)^2 * [sin (theta)]^2 / 2g

    but explanation and proving is still the same~ :)

  3. John

    the greatest sin value of theta is the sin of 45*

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