a stone is thrown in such a manner that it would just hit a bird at the top of a tree and afterwards reach a maximum height double that of the tree . if at the moment of throwing the stone the bird flies away horizontally with constant velocity and the stone hits the bird after some time . the ratio of horizontal velocity of stone to that of the bird is 1¡n+1¡¡Ìn?

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  1. Well, to make it easy, let's say the tree is 50 meters high. Then the max stone height will be 100 meters.
    That gives us a vertical problem to solve.
    What is Vi, initial vertical velocity?
    (1/2) m Vi^2 = m g h
    Vi = sqrt(2 g h = sqrt (200*9.81) = 44.3 m/s
    so when will it get to the tree top?
    h = Vi t - 4.9 t^2
    50 = 44.3 t - 4.9 t^2
    4.9 t^2 -44 t + 50 = 0
    t = [ 44 +/- sqrt(1936 -980) ] /9.81
    t = [ 44 +/- 30.9 ] /9.81
    t = 1.34 seconds on the way up, WHERE IT WOULD HAVE HIT THE BIRD
    t = 7.64 seconds on the way down - WHEN IT HITS THE BIRD!
    the bird flew for 7.64 seconds at speed u so went 7.64 u meters from the tree
    The stone went at horizontal speed s so went a total of 7.64 s meters
    of which 1.34 s was the distance from the thrower to the tree. so (7.64-1.34) s = 7.64 u
    I guess you can find s/u now :)

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