A hawk is flying horizontally at 10.0 m / s in a straight...

A hawk is flying horizontally at $10.0 \mathrm{~m} / \mathrm{s}$ in a straight line, $200 \mathrm{~m}$ above the ground. A mouse it has been carrying struggles free from its grasp. The hawk continues on its path at the same speed for $2.00 \mathrm{~s}$ before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse $3.00 \mathrm{~m}$ above the ground. (a) Assuming no air resistance, find the diving speed of the hawk. (b) What angle did the hawk make with the horizontal during its descent? (c) For how long did the mouse "enjoy" free fall?

A hawk is flying horizontally at $10.0 \mathrm{~m} / \mathrm{s}$ in a straight line, $200 \mathrm{~m}$ above the ground. A mouse it has been carrying struggles free from its grasp. The hawk continues on its path at the same speed for $2.00 \mathrm{~s}$ before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse $3.00 \mathrm{~m}$ above the ground. (a) Assuming no air resistance, find the diving speed of the hawk.
(b) What angle did the hawk make with the horizontal during its descent? (c) For how long did the mouse "enjoy" free fall?

Key Concepts

Vector Decomposition

Vector decomposition involves breaking a vector quantity into its horizontal and vertical components. This approach is crucial when analyzing motions in two dimensions, such as the hawk’s dive, where the overall velocity is resolved into components to calculate its magnitude and direction based on the contributions from each axis.

Kinematics Equations

These are the mathematical expressions that relate displacement, velocity, acceleration, and time. In problems involving projectile motion and free fall, the kinematic equations are used separately for horizontal and vertical motions to determine quantities like speed, angle, and duration of motion.

Projectile Motion

This concept involves the motion of objects that are launched into the air and continue moving under the influence of gravity alone, neglecting air resistance. It requires analyzing the motion in two independent components: horizontal motion, which occurs at constant velocity, and vertical motion, which is influenced by gravitational acceleration.

Free Fall

Free fall refers to the motion of an object when the only force acting upon it is gravity. In such cases, the object experiences a constant acceleration downward, typically approximated as 9.8 m/s² near the surface of the Earth. This is essential for calculating the time it takes for an object to fall a given vertical distance.

Transcript

0:00 Hello.

00:02 In this problem, we have a hoax that is 200 meters above the ground where it's holding a mouse.

00:11 At the first instant, the mouse escaped from the hoke and started to fall towards the ground.

00:20 And after two seconds, the hook kept moving in its direction.

00:26 So here, this is after two seconds, where the mouse kept falling.

00:31 And then the hook noticed this and started to move towards the mouse and she caught him where he was just three meters above the ground.

00:47 Of course the mouse kept moving in a projectile motion until it reached this position.

00:54 So if the hook moved in a straight line, what is the speed and the direction of the hook in this case? so let's start our analysis first by knowing after two seconds what happened.

01:19 So after two seconds, let's start.

01:25 After two seconds, the mouse's velocity in the x direction, ignoring the air resistance is just 10 meters per second as exactly the velocity of the holder, which is the hook.

01:40 And his y velocity is unknown at this point, and his height is unknown as well.

01:53 So let's get the height.

01:55 The height is always v .t.

01:57 T plus half a t squared.

02:01 In that case, the initial velocity in the y direction is zero, because he simply released himself, and a here stands for the gravitational acceleration, which is 9.

02:14 0 .8 times t squared.

02:17 T squared here we can replace it by 2 so that's 2 squared which is 4 so that's equal to 19 .6 meters downward from where he were originally so that this distance is 19 .6.

02:42 Starting from here let's know after what time did the mouse reach 3 meters above above the ground.

02:56 So again, let's substitute in that equation, but we put y here, the distance.

03:06 It needs to cross until it reach this height is 197 meters because we're calculating from 200 meters already.

03:17 So here we have 197 is equal to v .t again, v .0 is 0.

03:24 So this is equal to half a t squared.

03:28 So t is basically the square root of 197 divided by half multiplied by the gravitational acceleration, and that's 6 .3 .4 seconds...

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