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?
1 s

(b) How much additional time would be required for the fish's speedto triple again?
2 s
(a) How much time passes before the fish'sspeed triples?
1 s

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

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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