Question

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 4.50 m/s at an angle 35Á 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.

Answer 1

To determine if the mouse will fall into the nest, we need to calculate the horizontal position of the mouse when it has fallen 12.0 m. By using the equations of motion, we can find that the mouse will travel approximately 6.93 meters horizontally before hitting the ground.

Step-by-step explanation:

To determine if the mouse will fall into the nest, we need to calculate the horizontal position of the mouse when it has fallen 12.0 m.

First, we can calculate the time it takes for the mouse to fall 12.0 m using the equation h = (1/2)gt^2, where h is the initial vertical height, g is the acceleration due to gravity, and t is the time. Plugging in the values h = 12.0 m and g = 9.8 m/s^2, we can solve for t to find that it takes approximately 1.54 seconds for the mouse to fall 12.0 m.

Next, we can calculate the horizontal distance the mouse travels during this time using the equation d = vt, where d is the horizontal distance, v is the horizontal velocity, and t is the time. Plugging in the values v = 4.50 m/s and t = 1.54 seconds, we can solve for d to find that the mouse will travel approximately 6.93 meters horizontally before hitting the ground.

Answer 2

Actually it doesn't fall into the nest
Vertical component of the velocity, v of mouse after travelling vertical distance of 12 m is given by
v^2 - (4*sin 35)^2 = 2*9.81*12 or v = 15.51 m/s
time taken to travel 12 m downwards = (15.51 - 4*sin 35)/9.81 = 1.35 s
Horizontal distance traveled in this time = 4*(cos 35)*1.35 = 4.41 m
But the eastern end of the nest is only 4.15 m away