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A man of height 1.8m walks away from a lamp at a height of 6m. If the man's speed is 7m/s, find the speed in m/s at which the tip of the shadow moves.

  • physics -

    draw a right triangle: vertical 6m, horizontal leg x

    Now within that triangle, another vertical 1.8 m, and the horizontal distance to the lamppost is L, so the length to the tip of shadow is x-L

    similar triangles:
    x/6=(x-L)/1.8
    or 1.8x-6x=-6L

    4.2 dx/dt=6dL/dt

    given: dL/dt=7, solve for dx/dt

  • physics -

    h=1.8 m, H=6 m, v=7 m/s u=?

    x is the distance of man to lamp post,
    y is the distance from the tip of the shadow to the lamp post.

    From the similar triangles
    H/h = y/(y-x),
    6/1.8 = y/(y-x),
    6y -6x=1.8y,
    4.2 y=6 x,
    0.7y=x,
    d{0.7y}/dt=dx/dt,
    0.7(dy/dt)=dx/dt,
    0.7u =v,
    u=v/0.7=7/0.7=10 m/s.

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