mechanical

n projectile motion if range is the maximum, then what will be the height?.

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asked by Avais
  1. recall that maximum range occurs at angle = 45 degrees.
    also, the max height in projectile motion is given by:
    y,max = (Vo)^2 * sin^2 (theta)/(2g)
    where
    Vo = initial velocity
    theta = angle of release
    g = acceleration due to gravity = 9.8 m/s^2

    note that max height occurs at theta = 90 because sin^2 (theta) = 1 ---(when the object is thrown upwards, and thus the range is zero):
    y,max = (Vo)^2 / 2g

    but at theta=45,
    y = (Vo)^2 * sin^2 (45) / 2g
    y = (Vo)^2 * (sqrt(2)/2)^2 / 2g
    y = 1/2 * [(Vo)^2 / 2g]
    y = 1/2 * y,max

    at 45 degree angle release, the height is only half of the maximum it can reach (which is at 90 degree angle release)

    hope this helps~ :)

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    posted by Jai

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