A fish is able to jump vertically out of the water with a speed of 4.45 m/s. What is the speed of the fish as it passes a point 0.6 m above the water? How much time, after leaving the water does it take for the fish to pass a point 0.6 m above the water while it is on its way down?

Explanation:

given

initial velocity u = 4.45m/s

Height = 0.6m

g = 9.8m/s²

Required

final velocity v

Using the equation of motion;

v² = u²-2gH (upward motion of the fish makes g to be negative)

v² = 4.45²-2(9.8)(0.6)

v² = 19.8025-11.76

v² = 8.0425

v = 2.84 m/s

Hence the speed of the fish as it passes a point 0.6 m above the water is 2.84m/s

To get the time, we will use the formula

v = u - gt

2.84 = 4.45 - 9.8t

2.84-4.45 = -9.8t

-1.61 = -9.8t

t = 1.61/9.8

t = 0.164secs

Hence the time taken is 0.164secs

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