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Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is 37.5 m/s and that the second arrow is fired 1.66 s after the first. Determine the initial speed of the second arrow.

  • physics -

    The first arrow moved during the time interval ‘t’ to its highest point where its velocity is zero
    v=v1-g•t1, v=0 =>
    t1=v1/g =37.5/9.8 =3.83 s.
    The second arrow moves during t2= 3.83-1.66 =2.17 s.
    v =v2 - g•t2, v=0 =>
    v2= g•t2=9.8•2.17 =21.27 m/s.

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