An owl is carrying a mouse to the chicks in its nest. Its position at that
time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter
nest. The owl is flying east at 3.50 m/s at an angle
30.0º
below the
horizontal when it accidentally drops the mouse. Is the owl lucky enough
to have the mouse hit the nest? To answer this question, calculate the
horizontal position of the mouse when it has fallen 12.0 m
im struggling with this one as well. i got 0.23m.
To determine whether the mouse will hit the nest, we need to calculate its horizontal position when it has fallen 12.0 m.
First, let's break down the given information:
- The owl's position at that time is 4.00 m west and 12.0 m above the center of the nest.
- The owl is flying east at 3.50 m/s.
- The owl is flying at an angle of 30.0º below the horizontal when it drops the mouse.
- We need to calculate the horizontal position of the mouse when it has fallen 12.0 m.
To solve this problem, we'll use the following steps:
Step 1: Determine the time it takes for the mouse to fall 12.0 m.
We can use the formula for vertical motion: y = vit + 1/2at^2, where
- y is the vertical displacement (12.0 m),
- vi is the initial vertical velocity (0 m/s, as the mouse is dropped),
- a is the acceleration due to gravity (-9.8 m/s^2),
- t is the time taken.
Plugging in the given values, we have:
12.0 = 0*t + (1/2)*(-9.8)*t^2
Simplifying the equation, we get:
4.9t^2 = 12.0
Solving for t, we find:
t^2 = 12.0/4.9
t ≈ 1.24 s
Step 2: Calculate the horizontal distance traveled by the mouse in 1.24 seconds.
Since the owl is flying at a constant velocity of 3.50 m/s horizontally, we can multiply this velocity by the time taken (1.24 s) to find the horizontal distance traveled by the owl.
Horizontal distance = velocity * time
Horizontal distance = 3.50 m/s * 1.24 s
Horizontal distance ≈ 4.34 m
Step 3: Determine the horizontal position of the mouse relative to the center of the nest.
The owl starts 4.00 m west of the nest, and the horizontal distance traveled by the mouse is 4.34 m. Therefore, the absolute horizontal position of the mouse will be 4.00 m + 4.34 m = 8.34 m east of the nest.
Finally, to answer the question: Since the owl drops the mouse approximately 8.34 m east of the nest, and the nest has a diameter of 30.0 cm (or 0.30 m), the mouse will not hit the nest. It will fall beyond the edge of the nest.
To determine if the mouse will hit the nest, we need to calculate the horizontal position of the mouse when it has fallen 12.0 m.
Let's break down the given information:
- The owl's initial position is 4.00 m west and 12.0 m above the center of the nest.
- The owl is flying east at 3.50 m/s at an angle 30.0º below the horizontal.
To find the horizontal position of the mouse, we need to calculate the time it takes for the mouse to fall 12.0 m and then multiply it by the horizontal velocity.
Step 1: Calculate the time it takes for the mouse to fall 12.0 m
We can use the equation of motion: h = (1/2) * g * t^2, where h is the height, g is the acceleration due to gravity, and t is the time.
Given: h = 12.0 m, g = 9.8 m/s^2
Rearranging the equation, we have:
t^2 = (2h) / g
t^2 = (2 * 12.0) / 9.8
t^2 = 24.0 / 9.8
t^2 ≈ 2.45
Taking the square root of both sides, we find:
t ≈ √2.45
t ≈ 1.57 seconds
Step 2: Calculate the horizontal position of the mouse
The horizontal position is determined by multiplying the horizontal velocity by the time taken (t).
Given: horizontal velocity = 3.50 m/s
Horizontal position = horizontal velocity * time
Horizontal position = 3.50 m/s * 1.57 s
Horizontal position ≈ 5.50 m
Therefore, the mouse will miss the nest since its horizontal position is 5.50 m, which is 50 cm away from the center of the nest.