Horizontal Distance of Hammer Dropped from Roof: Solved

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Discussion Overview

The discussion revolves around calculating the horizontal distance a hammer travels after being dropped from a roof that is inclined at an angle of 30 degrees. Participants explore the implications of the angle of the roof on the components of velocity and the resulting motion equations, addressing both conceptual and mathematical aspects of the problem.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially assumes the angle of the hammer's trajectory is 60 degrees, leading to calculations based on this assumption.
  • Another participant suggests using the angle of 30 degrees for the horizontal component of velocity, indicating a potential misunderstanding of the angle's reference.
  • A clarification is provided that the angle should be considered with respect to the horizontal, emphasizing the importance of using the correct trigonometric functions based on the chosen angle.
  • Concerns are raised about the signs of the velocity components, particularly in relation to the direction of motion and the resulting quadratic equation.
  • Some participants express confusion about which angle to use and how it affects the calculations, highlighting the need for clarity in trigonometric applications.
  • There is a suggestion to draw the problem to better visualize the scenario and understand the relationships between the angles and the components of motion.
  • A later reply mentions that consistency in the choice of signs for the velocity components is crucial, regardless of the direction assumed.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct angle to use for the calculations, with multiple viewpoints on whether to use 30 degrees or 60 degrees. The discussion remains unresolved regarding the implications of these choices on the final calculations.

Contextual Notes

Participants note potential issues with negative values under the radical in the quadratic equation, suggesting that this may stem from misinterpretations of initial conditions and sign conventions.

Who May Find This Useful

This discussion may be useful for students and individuals interested in understanding the application of trigonometry in physics problems, particularly in the context of projectile motion and inclined planes.

Beretta
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A worker on the roof of a house drops her hammer which slides down the roof at constant speed of 4m/s. Th roof makes an angle of 30 with the horizontal, and the lowest point is 10m from the ground. what is the the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground?

What happen here is that I assumed that theta is 60 since it makes an angle if 30 with the horizontal and since the object is sliding down. Thus it leaves the roof on a 60 degree angle.

vx = 4m/s cos 60 = 2m/s
vy = 4m/s sin 60 = 3.46m/s

x(t) = x0 + vxt

y(t) = y0 + v0t - (1/2)g(t^2)
10m - 3.46m/s(t) + (0.5)(9.81m/s^2)t^2 = 0

I am ending up with a negative root Delta.
May you help me pls?
 
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Motion in 2 dimensions

try:
vx = 4 cos 30
vy = 4 sin 30
 
Why would the angle be 30 degree not 60 degree?
 
What you want is the angle that the hammer is moving with respect to the horizontal. THe hammer travels the same plane as the roof, which is sloped 30 degrees to the horizontal. THe 60 degrees is the angle made with a vertical line. IF you use the angle from vertical, you have to switch your sines and cosines.

You need to find the time of travel using your vertical information. It will send you into a quadratic solution. If you are getting a negative under the radical, it's probably because you are not recognizing that both acceleration and y-displacement are negative values (since you are calling initial y-velocity negative).
 
Should I leave it 30 then? I am really confused!
 
Try drawing it out. You can use either 30 or 60, but you must remember to use the right trignometric function to find the appropriate component.
 
Motion in two dimensions

Chi Meson and Moose are right. You should start this out by drawing the problem(always!). Watch your signs, you are in charge so you get to decide which direction is negative, however, you also have to answer to your instructor.
The important thing to be learned from this problem is that the speed is proportional to the length of the sides of the right triangle. Please remember this, think about it, and use it often so that you don't forget it.
Trigonometry can be very confusing at first, but if you bear this one thing in mind it will help you greatly; c^2 = a^2 + b^2.
Everything else in trigonometry is just a restatement of this.
To solve your problem you could use:
vx = 4 cos 30
vy = 4 sin 30
or
vx = 4 sin 60
vy = 4 cos 60
The result will be the same either way. It's not just a good idea, it's the law.
 
I am really greatful to everyone. Thank you all. One more question though, shouldn't be vx = -4 cos 30 and vy = -4 sin 30 since the object is sliding in the negative direction?
 
Anything you want, as long as you are consistent. Generally, left negative, right positive is the traditional system.
 

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