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An owl is carrying a mouse to the chicks in its nest. It is 4.00 m west and 12.0 m above the center of the 30 cm diameter nest and is flying east at 2.50 m/s at an angle 32° below the horizontal when it accidentally drops the mouse. Will it fall into the nest? Find out by solving for the horizontal position of the mouse (measured from the point of release) when it has fallen the 12.0 m.

Respuesta :

Answer:

4.61 m away

Explanation:

No, the owl will not fall into the nest.

The vertical component of the velocity, v of mouse after travelling vertical distance of 12 m is given by

v² - (4 * sin 32)² = 2 * 9.81 * 12

v² - (4 * 0.53)² = 235.44

v² - 4.494 = 235.44

v² = 235.44 + 4.494

v² = 239.934

v = 15.49 m/s

The time taken to travel the said 12 m downwards =

(15.49 - 4 * sin 32) / 9.81

(15.49 - 4 * 0.53) / 9.81

(15.49 - 2.12) / 9.81

13.37 / 9.81 = 1.36 s

Horizontal distance traveled in this time = 4 * (cos 32) * 1.36

= 4 * 0.848 * 1.36

= 4.61 m

Therefore, the horizontal position of the mouse is 4.61 m away from the point of release.

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