physics mechanic


(a) A rescue plane wants to drop supplies to isolated mountain climbers on a rocky range ridge 200m below. If the plane is travelling horizontally with a speed of 80ms-1, how far in advance of the recipients must the goods be dropped?

(b) Suppose, instead, that the plane releases the supplies a horizontal distance of 400m in advance of the mountain climbers. What vertical velocity should they give the supplies so that they arrive precisely at the climbers' position? With what speed do the supplies land in this case? Comment on the results.

( Ans: (a) 506m (b) -15ms-1, 113.6ms-1 )

1 Answer

  • 1 decade ago
    Favorite Answer

    (a) In vertical motion part, its time of flight can be found as:

    s = ut + at2/2

    with s = 200, u = 0 and a = 10

    t = 6.32 s

    So, in horizontal direction, it has flied for 6.32 x 80 = 506 m

    (b) Now the time of flight is 400/80 = 5 s

    So in vertical direction, it should fall a distance of 200 m in 5 s:

    s = ut + at2/2

    with s = 200, t = 5 and a = 10

    200 = 5u + 125

    u = 15 m/s

    Also, it vertical velocity of landing is:

    v2 - u2 = 2as with u= 15, a = 10 and s = 200

    v = 65 m/s

    In addition to its horizontal velocity of 80 m/s, by Pyth thm, the resultant velocity is 113.6 m/s

    Source(s): Myself
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