Calculate Components of Hammer's Velocity at Edge of Roof

In summary, a hammer slides down a roof at a 32.0 degree angle with the horizontal. The hammer's velocity at the edge of the roof is 6.25m/s. The magnitude of the horizontal velocity component is 5.21m/s and the magnitude of the vertical downward velocity component is 3.44m/s. The angles are not relevant in calculating the components of the hammer's velocity.
  • #1
murfzee0224
6
0
1. A hammer slides down a roof that makes a 32.0 degree angle with the horizontal.

2. a.) What are the magnitudes of the components of the hammer's velocity at the edge of the roof if it is moving at a speed of 6.25 meters per second?
b.) Calculate the components of the hammer's velocity at the edge of the roof.

3. I drew a picture for this problem and the teacher said it was right, but I don't know what to do from there. I ended up forcing numbers and have current answers of Vy = 6.25 sin 32.0 (3.44 m/s) with a magnitude of 32 degrees and Vx = 6.25 cos 32.0 (5.21 m/s) with a magnitude of 58 degrees.

PLEASE HELP ME! I know that what I have is wrong. I have checked my notes, the book, and classwork for a problem similar to this and I've had no luck. The teacher will not give anymore hints and my classmates haven't made any progress. Any help is appreciated, thanks!
 
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  • #2
murfzee0224 said:
1. A hammer slides down a roof that makes a 32.0 degree angle with the horizontal.

2. a.) What are the magnitudes of the components of the hammer's velocity at the edge of the roof if it is moving at a speed of 6.25 meters per second?
b.) Calculate the components of the hammer's velocity at the edge of the roof.

3. I drew a picture for this problem and the teacher said it was right, but I don't know what to do from there. I ended up forcing numbers and have current answers of Vy = 6.25 sin 32.0 (3.44 m/s) with a magnitude of 32 degrees and Vx = 6.25 cos 32.0 (5.21 m/s) with a magnitude of 58 degrees.

PLEASE HELP ME! I know that what I have is wrong. I have checked my notes, the book, and classwork for a problem similar to this and I've had no luck. The teacher will not give anymore hints and my classmates haven't made any progress. Any help is appreciated, thanks!
Yes, the hammer has a horizontal velocity component of 5.21m/s and a vertical downward velocity component of 3.44m/s. You have calculated those components correctly based on the given angle; why are you including angles in your calculated velocity components?
 
  • #3
Thanks for the help. I included the angles because I thought that they may have been the components' magnitudes.
 

1. How is the velocity of the hammer at the edge of the roof calculated?

The velocity of the hammer at the edge of the roof can be calculated using the formula: V = √(2gh), where V is the velocity, g is the gravitational acceleration (9.8 m/s^2), and h is the height of the roof.

2. What are the components of the hammer's velocity at the edge of the roof?

The components of the hammer's velocity at the edge of the roof are the horizontal and vertical components. The horizontal component is equal to the initial horizontal velocity of the hammer, while the vertical component is equal to the final vertical velocity of the hammer.

3. How does the height of the roof affect the components of the hammer's velocity?

The height of the roof directly affects the vertical component of the hammer's velocity. As the height increases, the vertical component of the velocity also increases. The horizontal component remains constant regardless of the height of the roof.

4. Can the components of the hammer's velocity be negative?

Yes, the components of the hammer's velocity can be negative. The negative sign indicates the direction of the velocity, where a negative horizontal component means the hammer is moving towards the left and a negative vertical component means the hammer is moving downwards.

5. How can the components of the hammer's velocity be used in real-world situations?

The components of the hammer's velocity can be used to determine the trajectory of the hammer after it is thrown off the roof. This information can be useful in predicting where the hammer will land and ensuring the safety of people and objects in the surrounding area.

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