SOLUTION: On Christmas day 2007 a tiger leapt from its enclosure at San Francisco Zoo. Official guidelines said a wall to contain tiger's should be 5 metres high. The zoo's wall was only 3.8

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Question 986260: On Christmas day 2007 a tiger leapt from its enclosure at San Francisco Zoo. Official guidelines said a wall to contain tiger's should be 5 metres high. The zoo's wall was only 3.8 metres high but the tiger also had to jump a 10 metre moat. The tiger could run at the moat before leaping at the wall (curved path) . With a 5 metre run-up, tigers reach 14 metres per second at take-off. This formula calculates how fast an animal has to be moving to be able to leap over a moat AND a wall.
V=sqrt(9.8(h+sqrt(h^2+w^2))
where V=takeoff speed, h=height of wall w=width of moat.
i) Would the tiger have been able to leap over a wall 5 metres high if I was running at 14 metres per second? (You must support your answer with calculations using the formula).
ii) If it wasn't possible to make the wall higher than 3.8 metres, then you could make the moat wider. By rearranging the equation to make w the subject calculate how wide a moat is needed to contain a tiger running at 14 metres per second?
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
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V=takeoff speed; h=height of wall; w=width of moat.
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V=sqrt%289.8%28h%2Bsqrt%28h%5E2%2Bw%5E2%29%29%29
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1) Would the tiger have been able to leap over a wall 5 meters high if it was running at 14 meters per second? (moat=w=10 meters)
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14=sqrt%289.8%28h%2Bsqrt%28h%5E2%2B10%5E2%29%29%29
196=9.8h%2B9.8sqrt%28h%5E2%2B100%29
20=h%2Bsqrt%28h%5E2%2B100%29
20-h=sqrt%28h%5E2%2B100%29 Square both sides
%2820-h%29%5E2=h%5E2%2B100
h%5E2-40h%2B400=h%5E2%2B100
-40h%2B300=0
-40h=-300
h=7.5meters
ANSWER: Running at 14 m/s, the tiger could jump over a 10 meter moat and a wall 7.5 meters high.
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2) How wide must the moat be to contain a tiger running at 14 meters per second?
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V=sqrt%289.8%28h%2Bsqrt%28h%5E2%2Bw%5E2%29%29%29
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V%5E2=9.8%28h%2Bsqrt%28h%5E2%2Bw%5E2%29%29
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V%5E2%2F9.8=h%2Bsqrt%28h%5E2%2Bw%5E2%29
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%28V%5E2%2F9.8%29-h=sqrt%28h%5E2%2Bw%5E2%29
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%28%28V%5E2%2F9.8%29-h%29%5E2=h%5E2%2Bw%5E2
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%28%28V%5E2%2F9.8%29-h%29%5E2-h%5E2=w%5E2
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sqrt%28%28%28V%5E2%2F9.8%29-h%29%5E2-h%5E2%29=w
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For V=14 m/s and h=3.8 m:
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sqrt%28%28%28V%5E2%2F9.8%29-h%29%5E2-h%5E2%29=w
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sqrt%28%28%2814%5E2%2F9.8%29-3.8%29%5E2-3.8%5E2%29=w
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sqrt%2816.2%5E2-3.8%5E2%29=w
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sqrt%28248%29=w
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12sqrt%282%29=w or approximately 16.97 meters
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ANSWER: With a 3.8 meter wall, the moat must be 16.97 meters wide to contain the tiger.