# Calculating the speed (2d motion)

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This is a question from a book that I'm reading to learn Physics:
Quote:
 When the sun is directly overhead, a hawk dives toward the ground with a constant velocity of 5.00 m/s at 60.0 degrees below the horizontal. Calculate the speed of the hawks shadow on the level ground.
Here is what I think the graph looks like:
     |
_____|________________
|\
| \  @ = 60.0
|  \
|   
I know that the speed is the magnitude of the vector, but my problem is finding the vector components or maybe there is another vector that I should be calculating. I know that the vector components are: Vx = V cos(@) Vy = V sin(@) Would this mean that: Vx = 2.5 m/s Vy = -4.330127019 m/s I wanted to make sure that Vx and Vy are correct so I did: tan @ = Vy/Vx which equals the angle the hawk is fling in, so I know the V components are correct. One guess I thought was that the shadow would be moving at the speed of the Vx because its moving across the x axis, but I belive that is wrong although the answer the book says is 2.5 m/s. I just doesn't seem like that would be all I have to do to solve the problem. If I take the magnitude of V which is the speed: |V| = sqrt(Vx^2 + Vy^2) = 5 And that doesn't seem right. I'm thinking that I'm not going in the right direction with this problem. Am I missing something or not using an Equation or something. What do you think? Thanks you for all relpies, - LostSource [Edited by - LostSource on June 8, 2006 8:53:21 PM]

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Ummmm... yeah, the books is right, try tossing a ball straight up, and if the sun is at zenith than where will the shadow go?

But if the sun is at the horizon than the shadow will move, this might have caught you, never seen any post a school stuff here.

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Quote:
 Original post by LostSource...One guess I thought was that the shadow would be moving at the speed of the Vx because its moving across the x axis, but I believe that is wrong although the answer the book says is 2.5 m/s...

Why do you believe it is not VX? You said yourself that the shadow moves along the X axis.
Quote:
 Original post by LostSource...If I take the magnitude of V which is the speed:|V| = sqrt(Vx^2 + Vy^2) = 5And that doesn't seem right...

Umm, the magnitude is 5. How can that not seem right?

It's kind of funny that you found the answer, verified your work two different ways, and got the same result as the book, yet you still think that it's wrong!

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Quote:
Original post by JohnBolton
Quote:
 Original post by LostSource...One guess I thought was that the shadow would be moving at the speed of the Vx because its moving across the x axis, but I believe that is wrong although the answer the book says is 2.5 m/s...

Why do you believe it is not VX? You said yourself that the shadow moves along the X axis.

Yeah, how much sleep did you get?

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Haha, thanks guys for the help. Thats funny that I solved it and didn't realize that I was doing it right. Ive been lacking in the sleeping department lately, but for some reason I get tried but the second I lay down I'm wide awake.

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