An eagle is flying horizontally at 5.5 m/s with a fish in its claws. it accidentally drops the fish. (a) how much time passes before the fish's speed doubles? (b) how much would additional time be required for the speed to double again?

Answer

given,

eagle horizontal velocity = 5.5 m/s

resultant speed is given by

t = 0.971 s

b)

resultant speed is given by

t' = 1.86 s

therefore additional time is equal to = t' - t

                                                            = 1.86 - 0.971

                                                            = 0.889 s

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

think its the third option