VI. The Big Ol\' Dek. A Big Ol Duck flies in a straight line due south at a cons
ID: 1582730 • Letter: V
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VI. The Big Ol' Dek. A Big Ol Duck flies in a straight line due south at a constant speed of 20 m/s. A Quiet Young Human stands 40 meters below the passing duck. She wishes to hit the duclk when the duck is directly overhead. Right at the moment she starts making plans to pursue this wish, however, the duck is not only above her, but also some noticeable (but unknown) distance north of her. (SEE DIAGRAM! [also: see how vital diagrams are?] (And please note: NORTH is NOT the same as UP) The Quiet Young Human fires a shot straight up with an initial velocity of 30 m/s up Of course, after the shot is fired, it is in free fall... which means constant (Fill in the blank in the line above using what you discovered in lab 2.) For this problem and for ALL future homework and exam problems, you may assume that (near earth's surface), free fall acceleration is 10 ms downwards Half a second after the shot is fired, The Big Ol Duck instantly commences a constant n of 16 meters/sec forward (to the south Vo 20 m/s aas 16 m/s 40 m South Vo:30 m/s a) How far North of the QYH should the Big Ol Duck be at the moment the shot is fired, if b) Why are there two mathematically possible answers to the above question (a)? Are both c) What if the Big Of Duck had accelerated at the same constant rate but toward the North? the shot is to hit the Big Ol' Duck? mathematical "roots" physically meaningful? Can you find four mathematical roots" to the problem? Are there four physically meaningful answers to the problem?Explanation / Answer
- Free fall means constant acceleration.
a) Time taken by shot to reach at height of 40m is given by equation
40 = 30 t - 10 t2 / 2
t = 2/4 sec
Distance travelled by Duck in these times,
d = 20(1/2) + 20(1.5) + 16(1.5)2/2 ( for t =2 sec) ....1
= 20(1/2) + 20(3.5) + 16(3.5)2/2 ( for t = 4 sec) ......2
d = 58/178 m
Hence Duck is 58/178 m north of QYH
b) There are 2 answers as shot reaches the height of 40 m twice. One while going up second while coming down.
c) If acceleration of Duck is towards north, we have to take a=-16, in equation 1 and 2.
From equation1 , that is t = 2 sec, we get d = 22m.
But from equation 2 we get -ve value of d. This means shot can not hit duck after 4 seconds and reason is, Duck reaches back to it's position at t=0 , in less than 3 seconds. So it can not be above QYH after 4 seconds.
Hence in this case there is only one possible answer.
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