Answer:
28.1 m/s
Explanation:
[tex]u_x[/tex] = Initial velocity of the fish = 1.52 m/s
y = Height of the bird = 40 m
[tex]a_y[/tex] = Acceleration in y axis = [tex]9.81\ \text{m/s}^2[/tex]
[tex]u_y[/tex] = Initial velocity in y axis = 0
[tex]y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 40=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{40\times 2}{9.81}}\\\Rightarrow t=2.86\ \text{s}[/tex]
[tex]v_y=u_y+a_yt\\\Rightarrow v_y=0+9.81\times 2.86\\\Rightarrow v_y=28.057\ \text{m/s}[/tex]
The final velocity in x direction will remain the same as the initial velocity as there is no acceleration in the x direction [tex]u_x=v_x=1.52\ \text{m/s}[/tex]
Resultant velocity is given by
[tex]v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{1.52^2+28.057^2}\\\Rightarrow v=28.1\ \text{m/s}[/tex]
The fish is moving at a velocity of 28.1 m/s when it hits the water.
