An owl is carrying a mouse to the chicks in its nest its...

An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.10 m west and 5.8 m above the center of the 33.0 cm diameter nest. The owl is flying east at 5.00 m/s at an angle 30.0° below the horizontal when it accidentally drops the mouse. How far from the center of the nest does the mouse land? Does the mouse land inside the nest? (To answer this question, calculate the horizontal position of the mouse when it has fallen 5.8 m). Distance from center of nest = Land inside nest? Yes No

Transcript

00:01 So in this problem, we have an owl flying back to their nest with a bird, with a mouse.

00:08 So we need to start with an axis system, x and y.

00:13 And we know that the owl is flying downward at some 30 degree angle with an initial velocity of 5 meters per second.

00:28 So when the mouse is let go, it will fall, but continue accelerating down.

00:35 Due to gravity, so it will have some curved trajectory as opposed to if there was no gravity, we would continue straight.

00:46 So there is a nest on the ground.

00:52 There is a nest on the ground with some diameter of 33 centimeters or 0 .33 meters diameter, where that's the diameter symbol.

01:04 And we want to see where the mouse lands.

01:06 Does it land in the nest? does it land too far? does it land? to short.

01:15 We know that the owl is 5 .8 meters above and the nest is 4 .1 meters across to its center.

01:31 So in these problems of projectile motion, what we need to do is look at the x and y components separately so that an object falling under gravity will take the same time to land as this object, the mouse that is moving some distance to the side.

01:50 So we can solve this problem of falling and then use it to explain the trajectory of the mouse.

01:59 So just in the y direction with constant acceleration, the kinematic equation that we need to use is the position is equal to the final position is equal to the initial position, plus the initial speed in that direction times time, plus one half the acceleration times time squared.

02:19 So from our axis system, the final position is equal to 0 because we set our origin, so the final y value is y equals 0.

02:30 And then acceleration is negative g.

02:33 The acceleration due to gravity on earth, negative because it's in the downward direction compared to our positive axis that we set earlier.

02:43 And so where we have those to plug in, we need to solve this quadratic equation for t, which we can do.

02:52 Negative b plus minus the square root of b squared minus four times a times c which is just why not and all of this is over two times a which is g over two so from this we can start plugging in numbers the initial velocity in the we need to use trigonometry for.

03:27 It'll have some component v .0x and v .0y.

03:34 And if we project over, you see v .0y here, which from this triangle that we have would be the sine 30 degrees of v .0, or 5 meters per second.

03:52 So we know that v .0y is 5 sine 30 degrees, and it's all negative.

04:06 Same thing inside the radical square both terms...

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