. A bird is flying with a speed of 18.0 m/s over water when it accidentally drops a 2.00 kg fish. If the altitude of the bird is 5.40m and friction is disregarded, what is the speed of the fish when it hits the water?

Here's what y

Respuesta :

v^2 = u^2 + 2gr, where v - speed, u - initial speed=0, r - displacement (or height)
v^2 = 0 + 2*10*5.4
v = 10.2 m/s

Answer:

20.73 m/s

Explanation:

Defining as state 1 the moment before the drop and state 2 the moment when the fish hits the water. Then:

speed at state 1, v1 = 18.0 m/s

mass of the fish, m = 2.00 kg

height at state 1, h1 = 5.40 m

height at state 2, h2 = 0 m

speed at state 2, v2 = ? m/s

acceleration of gravity, g = 9.8 m/(s^2)

We know that the addition of kinetic energy (1/2*m*v^2) and potential energy (m*g*h) at both states must be equal. Then:

1/2*m*v1^2 + m*g*h1 = 1/2*m*v2^2 + m*g*h2

Replacing with data (units are omitted):

1/2*2*18^2 + 2*9.81*5.4 = 1/2*2*v2^2 + 2*9.81*0

430 = v2^2

v2 = √430

v2 = 20.73 m/s