Question 986260
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V=takeoff speed; h=height of wall; w=width of moat.
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{{{V=sqrt(9.8(h+sqrt(h^2+w^2)))}}}
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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(9.8(h+sqrt(h^2+10^2)))}}} 
{{{196=9.8h+9.8sqrt(h^2+100)}}}
{{{20=h+sqrt(h^2+100)}}}
{{{20-h=sqrt(h^2+100)}}} Square both sides
{{{(20-h)^2=h^2+100}}}
{{{h^2-40h+400=h^2+100}}}
{{{-40h+300=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(9.8(h+sqrt(h^2+w^2)))}}}
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{{{V^2=9.8(h+sqrt(h^2+w^2))}}}
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{{{V^2/9.8=h+sqrt(h^2+w^2)}}}
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{{{(V^2/9.8)-h=sqrt(h^2+w^2)}}}
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{{{((V^2/9.8)-h)^2=h^2+w^2}}}
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{{{((V^2/9.8)-h)^2-h^2=w^2}}}
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{{{sqrt(((V^2/9.8)-h)^2-h^2)=w}}}
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For V=14 m/s and h=3.8 m:
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{{{sqrt(((V^2/9.8)-h)^2-h^2)=w}}}
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{{{sqrt(((14^2/9.8)-3.8)^2-3.8^2)=w}}}
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{{{sqrt(16.2^2-3.8^2)=w}}}
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{{{sqrt(248)=w}}}
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{{{12sqrt(2)=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.