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physics

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An athlete executing a long jump, leaves the ground at a 30 degree angle with the ground and travels 8.90 meters horizontally in 1.1 seconds. What was the takeoff speed of the athlete along the diagonal?

  • physics - ,

    D = Xo*t = 8.90 m
    Xo * 1.1 = 8.9
    Xo = 8.9/1.1 = 8.09 m/s = Hor. component
    of initial velocity.

    Vo = Xo/cos A = 8.09/cos30=9.34 m/s[30o]
    = Initial velocity or take-off speed.

  • physics - ,

    km,

  • physics - ,

    Δy = 0, assuming cannonball falls back down to same level
    a = -g
    vi = vsinθ

    Δy = vi(t) + ½a(t)²
    0 = v(sinθ)(t) + ½(-g)(t)²
    0 = t(vsinθ - ½gt)

    Clearly, there are two points in time when the projectile is at ground level.
    t = 0, is the time when the projectile is launched.
    let 0 = vsinθ - ½gt to find out when it lands

    ½gt = vsinθ
    t = 2vsinθ/g

    This result can also be obtained by showing that the initial vertical speed is equal to the final vertical speed but opposite in direction:
    vf² = vi² + 2gΔy
    since change in height is 0,
    vf² = vi²
    vf = -vi, since the projectile moves up on the way up and down on the way down

    Using: a = (vf - vi)/t
    t = (vf - vi)/a
    t = (-vi - vi)/a
    t = -2vi/a
    t = -2vsinθ/-g
    t = 2vsinθ/g


    [Range]

    To determine range, analyse motion in the horizontal or x-direction:
    Δx = vt, where v is the constant horizontal speed of the projectile
    Δx = (vcosθ)(2vsinθ/g)
    Δx = v²2cosθsinθ/g

    Using the identity: sin(2θ) = 2cosθsinθ,

    Δx = v²sin(2θ)/g

    v is launch speed
    θ is launch angle
    g is acceleration due to gravity (9.8 m/s²)


    [Launch Speed]

    v = sqrt(Δxg/sin(2θ))

    Multiply by 1.06 and use new speed in distance equation

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