An eagle is flying horizontally at 5.0m/s with a fish in its claws. It accidenta

Question

An eagle is flying horizontally at 5.0m/s with a fish in its claws. It accidentally drops the fish. (a) How much time passes before the fish'sspeed triples?

(b) How much additional time would be required for the fish's speedto triple again?

(a) How much time passes before the fish'sspeed triples?

(b) How much additional time would be required for the fish's speedto triple again?

Explanation / Answer

An eagle isflying horizontally at 5.0 m/s with a fish in its claws. It accidentally drops the fish. (a) How much timepasses before the fish's speed triples?     {Speed} = v = Sqrt[(v_x)^2 + (v_y)^2 ]      ---->    v = Sqrt[ (5)^2 + (g*t)^2 ]      ---->    v = Sqrt[ (5)^2 + ((9.8)*t)^2 ]      {v = 3*(5) =15} ---->   Sqrt[ (5)^2 +((9.8)*t)^2 ] = 15      ---->    (5)^2 +((9.8)*t)^2 = (15)^2      ---->    t = Sqrt[ (15)^2 - (5)^2]/(9.8)      ---->   {Time To TripleSpeed} = 1.443 sec (b) How muchadditional time would be required for the fish's speed totripleagain?     {Speed} = v = Sqrt[(v_x)^2 + (v_y)^2 ]      ---->    v = Sqrt[ (5)^2 + (g*t)^2 ]      ---->    v = Sqrt[ (5)^2 + ((9.8)*t)^2 ]      {v =(3)*(3)*(5) = 45} ---->   Sqrt[ (5)^2 + ((9.8)*t)^2 ] = 45      ---->    (5)^2 +((9.8)*t)^2 = (45)^2      ---->    t = Sqrt[ (45)^2 - (5)^2]/(9.8)      ---->    t = 4.563 sec      ---->    {AdditionalTime To Triple Again}} = (4.563) - (1.443)      ---->   {Additional TimeTo Triple Again} = 3.12 sec .

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