What is the Impact Velocity of a Mouse Dropped by an Owl?

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Homework Help Overview

The problem involves a mouse dropped by an owl flying horizontally at a speed of 5 m/s from a height of 10 m. Participants are tasked with determining the mouse's impact velocity, including both magnitude and angle, while grappling with the two-dimensional nature of the motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the quadratic formula to find time and the impact velocity, questioning the appropriateness of this method for a two-dimensional problem.
  • There are attempts to separate horizontal and vertical components of motion, with some uncertainty about how to calculate the angle of impact.
  • Concerns are raised about the initial vertical velocity and how to derive it without an angle provided in the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have made attempts to solve for time and velocity components, while others are still uncertain about the correct approach to finding the angle of impact. Guidance has been offered regarding the separation of motion components and the setup of equations.

Contextual Notes

Participants note the absence of an initial angle, which complicates their ability to find time and vertical velocity components. There is a recognition that vertical and horizontal motions are independent, but this leads to further questions about how to proceed without complete information.

Istinkatmath
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Homework Statement


An owl accidentally drops a mouse it was carrying while flying horizontally at a speed of 5 m/s. The mouse drops to the ground 10m below. What is the mouse's impact velocity? Give the answer as a magnitude and clearly defined angle.

As of right now I'm not even sure how to approach this problem, if I had an angle to work with I think I could get through it but at the moment I'm bamboozled.

Thanks for any help :D
 
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All i know is that you have to use quadratic formula on -4.9t^2+5t+10 to get the time it took to get to the ground. then divide 10m by t to get the speed at impact. no idea how to get the angle. on second thought i think the angle is the slope of the tangent of the parabola at where it hits the x-axis on the right side but not sure.
 
this is 2 dimensional though, wouldn't using the quadratic formula like that indicate that the mouse was shot straight up in the air and then is falling back down? I think I have to split it into X and Y components, but I'm not sure how to do so.

Edit: Duh, vertical motion is independent of horizontal motion, so yea you're right. Still unsure how to get the angle out of this though.
 
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Hi Istinkatmath,

Istinkatmath said:
this is 2 dimensional though, wouldn't using the quadratic formula like that indicate that the mouse was shot straight up in the air and then is falling back down? I think I have to split it into X and Y components, but I'm not sure how to do so.

Edit: Duh, vertical motion is independent of horizontal motion, so yea you're right. Still unsure how to get the angle out of this though.

The impact velocity is the final velocity (right before it hits ground). How can you find the components of the final velocity? Do you see what to do with them?
 
i think that's my problem at the moment, i don't see what to do with the components.

Edit: thanks for helping and caring enough to respondI solved the quadratic equation and got my final velocity to be 4.933 m/s, now I just have to figure out what to do with it : /
 
Istinkatmath said:
i think that's my problem at the moment, i don't see what to do with the components.

Edit: thanks for helping and caring enough to respond


I solved the quadratic equation and got my final velocity to be 4.933 m/s, now I just have to figure out what to do with it : /

I don't see how you got that result. Can you show the work you did to get it?

Remember that the horizontal and vertical components are handled separately. So think about what you know for the horizontal motion (initial horizontal velocity, final horizontal velocity, horizontal displacement, horizontal acceleration and time). You're looking for the x-component of the final velocity, so how could you find it?

Separately, think about the vertical motion and what you know about it in this problem (initial vertical velocity, final vertical velocity, vertical displacement, vertical acceleration and time). With these, you're looking for the y-component of the final velocity.
 
yea i was wrong. so when i am setting up the y(t) function for motion, would it be the following? y= 10 + Voy(t)-1/2gt^2 ? All of the examples in my book give an angle to work with from the beginning, and I don't understand how I can find t without it.
 
Last edited:
Istinkatmath said:
yea i was wrong. so when i am setting up the y(t) function for motion, would it be the following? y= 10 + Voy(t)-1/2gt^2 ? All of the examples in my book give an angle to work with from the beginning, and I don't understand how I can find t without it.

That's right; the general equation is:

[tex] y=y_0 + v_{0y} t +\frac{1}{2}a_yt^2[/tex]

You've already plugged in [itex]y_0=10[/itex] and [itex]a_y=-g[/itex]. You also know (from the problem) what [itex]y[/itex] and [itex]v_{oy}[/itex].

The value [itex]y[/itex] is the coordinate of the mouse when it hits the ground; if it starts at 10m, what is it at after it falls 10m?

The value [itex]v_{0y}[/itex] is how much of the initial velocity that is in the vertical direction. What would that be?

Once you plug those in, your only unknown should be time and you can find it.
 

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