Question

I was having trouble with this problem, and problems like it: A 3.2 kg pelican, with a 1.73 kg fish in its mouth, is flying 1.52 m/s at a height of 40 m when the fish wiggles free and fall back toward the ocean. How fast is the fish moving when it hits the water?

Answer 1

Answer:

28.1 m/s

Explanation:

$u_x$ = Initial velocity of the fish = 1.52 m/s

y = Height of the bird = 40 m

$a_y$ = Acceleration in y axis = $9.81\ \text{m/s}^2$

$u_y$ = Initial velocity in y axis = 0

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 40=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{40\times 2}{9.81}}\\\Rightarrow t=2.86\ \text{s}$

$v_y=u_y+a_yt\\\Rightarrow v_y=0+9.81\times 2.86\\\Rightarrow v_y=28.057\ \text{m/s}$

The final velocity in x direction will remain the same as the initial velocity as there is no acceleration in the x direction $u_x=v_x=1.52\ \text{m/s}$

Resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{1.52^2+28.057^2}\\\Rightarrow v=28.1\ \text{m/s}$

The fish is moving at a velocity of 28.1 m/s when it hits the water.