Question

A 12.9 kg pelican, with a 1.51 kg fish in its mouth, is flying 14.94 m/s at a height of 114 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:

v =  49.6 m/s

Explanation:

The vertical component of the final speed can be found by using 3rd equation of motion:

$2gh = v_{y}^{2} - v_{0y}^{2}$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 114 m

Voy = vetical component of initial velocity = 0 m/s (since the pelican is flying horizontally)

Vy = Vertical component of final velocity = ?

Therefore,

$(2)(9.81\ m/s^{2})(114\ m) = v_{y}^{2} - (0\ m/s)^{2}\\v_{y} = \sqrt{2236.68\ m^{2}/s^{2}}\\v_{y} = 47.3\ m/s$

Now, the horizontal component of the velocity is assumed to be the same because of negligible frictional force. Therefore,

Vx = 14.94 m/s

Thus the final speed will be given as:

$v = \sqrt{v_{x}^{2} + v_{y}^{2}}\\\\v = \sqrt{(14.94\ m/s)^{2} + (47.3\ m/s)^{2}}$

v =  49.6 m/s