(a) The time before the speed doubles is 0.71 s
(b) The additional time required is 1.43 s
The given parameters;
velocity of the eagle, v = 7 m/s
the velocity of the fish relative to the ground = 7 m/s
The time when the speed doubles is calculated as follows;
[tex]v_f ^2 = v_o^2 + 2gh[/tex]
where;
h is the height traveled by the fish when the speed doubles;
Vf = 2 x Vo
[tex](14)^2 = 7^2 + 2\times 9.8h\\\\196 = 49 + 19.6h\\\\19.6h = 196-49\\\\19.6h = 147\\\\h = \frac{147}{19.6} = 7.5 \ m[/tex]
The time when the distance traveled is 7.5 m;
[tex]h = v_ot + \frac{1}{2} gt^2\\\\7.5 = 7t + \frac{1}{2} \times 9.8 t^2\\\\7.5 = 7t + 4.9t^2\\\\4.9t^2 + 7t - 7.5 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method\\\\a = 4.9, \ b = 7, \ c = -7.5\\\\t = \frac{-b \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(7) \ +/- \ \ \sqrt{(7)^2 - 4(4.9\times -7.5)} }{2(4.9)} \\\\t = 0.71 \ s[/tex]
Thus, the time before the speed doubles is 0.71 s
(b) The additional time required when the speed double again;
[tex]v_f = 2\times 14 = 28 \ m/s\\\\v_f^2 = v_o^2 + 2gh\\\\28^2 = 14^2 + 2\times 9.8h\\\\784 = 196 + 19.6h\\\\h = 30 \ m\\[/tex]
The time of motion when the distance is 30 m;
[tex]30 = 14t + 0.5\times 9.8 t^2\\\\30 = 14t + 4.9t^2\\\\4.9t^2 + 14 t - 30 = 0\\\\a = 4.9 , b = 14, \ c = -30\\\\t = \frac{-b \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\ t = \frac{-(14) \ +/- \ \ \sqrt{(14)^2 - 4(4.9\times -30)} }{2(4.9)} \\\\t= 1.43 \ s[/tex]
Thus, the additional time required is 1.43 s
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