physics

posted by mona

A body is projected vertically upwards with a speed of 40 m/s.Find the distance travelled by body in last second of upward journey?(g=9.8m/s2 neglect effect of air resistance)

  1. Henry

    V = Vo + g*Tr = 0 at max. ht.
    Tr = -Vo/g = -40/-9.8 = 4.08 s. = Rise
    time.

    T = 4.08-1 = 3.08 s.
    h1 = Vo*t + 0.5g*t^2
    h1 = 40*3.08 - 4.9*3.08^2 = 76.7 m.

    h2 = 40*4.08 - 4.9*4.08^2 = 81.6 m.

    h2-h1 = 81.6-76.7 = 4.93 m. = Distance traveled in the last second.

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