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
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