physics

A projectile hits a target placed on a level field. It has arange of 37.0 m. The nagle of launch is 33.4 degrees above the horizontal. What was the initial speed?

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asked by Anne
  1. The horizontal component of velocity was constant so
    (V cos 33.4)t = 37
    V t = 44.3
    V = 44.3/t

    in the up direction
    y = 0 + V sin 33.4 t - 4.9 t^2
    0 = 0 + 24.38 =4.9 t^2
    4.9 t^2 = 24.38
    t^2 = 4.98
    t = 2.23seconds in air
    V = 44.3/2.23
    V = 19.9 m/s

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    posted by Damon

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