physics

an arrow is shot at a 30 degree angle with the horizontal. it has a velocity of 49 m/s how high will the arrow go? what horizontal distance will the arrow travel? and how long will the arrow be in the air?

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  1. Vertical problem:
    Vi = 49 sin 30 = 24.5 m/s
    v = Vi - 9.81 t
    at top v = 0
    0 = 24.5 - 9.81 t
    t = 2.50 seconds upward
    so time in air = 2*2.5 = 5 seconds

    h = Vi t - 4.9 t^2
    at t 2.5
    h = 24.5*2.5 - 4.9(6.25)
    = 61.25 - 30.625
    = 30.625 meters high

    Horizontal:
    u = 49 cos 30 forever or
    at least until the ground hits it.
    so
    range = 49 cos 30 * 5 seconds

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  2. Vo = 49m/s[30o] = Initial velocity.
    Xo = 49*Cos30 = 42.4 m/s = Ver. component.
    Yo = 49*sin30 = 24.5 m/s = Hor. component.

    A. Y^2 = Yo^2 + 2g*h.
    0 = 24.5^2 + (-19.6)*h,
    h = 30.63 meters.

    B. Range = Vo^2*sin(2A)/g
    Range = 49^2*sin(60)/9.8 = 212.2 m.

    C. Y = Yo + g*Tr.
    0 = 24.5 + (-9.8)Tr,
    Tr = 2.5 s. = Rise time.
    Tf = Tr = 2.5 s = Fall time.
    Tr+Tf = 2.5 + 2.5 = 5.0 s. = Time in air.

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