The mouse will be 4.61 meters distant horizontally. and the owl won't fall into the nest.
The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
Diameter of nest,d = 30 cm
The velocity of flying, V = 3.50 m/s toward the east
The angle of flying below the horizontal, Θ = 32°
Position of owl,12.0 m above the middle of the 30 cm-diameter nests, at 4.00 m west.
Height of owl above the ground,h = 12 m
Distance from the west direction, a = 4.00 m
The vertical component of velocity,vₐ
The horizontal component of velocity,vₓ
After covering a vertical distance of 12 m, the vertical component of the mouse's velocity, v, is given by
vₐ-u²=2gh
u = asinΘ
Substitute the above value;
vₐ²- (a sin Θ )² = 2gh
Substitute the given value;
vₐ² - (4 sin 32)² = 2 ×9.81 ×12
vₐ² - (4 × 0.53)² = 235.44
vₐ² - 4.494 = 235.44
vₐ² = 235.44 + 4.494
vₐ² = 239.934
vₐ = 15.49 m/s
The time required to descend the aforementioned 12 meters;
[tex]\rm t = \frac{v_{rel}}{g}[/tex]
t = (vₐ - u )/g
t =( vₐ - a sin Θ )/g
Substitute the given value;
t =(15.49 - 4 ×sin 32°) / 9.81
t= (15.49 - 4 * 0.53) / 9.81
t =(15.49 - 2.12) / 9.81
t = 13.37 / 9.81
t= 1.36 s
Horizontal distance traveled for the given period;
x=uₓ × t
x =a cosΘ × t
Substitute the given value;
x = 4×(cos 32°)×1.36
x= 4×0.848×1.36
x= 4.61 m
Hence, the mouse will be 4.61 meters distant in the horizontal direction. and the owl won't fall into the nest,
To learn more about the velocity, refer to the link: https://brainly.com/question/862972.
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