text: # Projectile Motion - hammer sliding off a roof
  
  1. Sep 17, 2011
  
  ### Nickg140143
  
  1. The problem statement, all variables and given/known data
  A hammer slides off the roof of a house that slopes downward at 40 degrees. The edge of teh roof is 14m avove the ground and the hammer leaves the roof at 7m/s.
  
  How far from the edge of the roof does the hammer strike the ground?
  
  2. Relevant equations
  x direction
  Vx=V0x
  
  x=V0cosθ
  
  y direction
  Vy=V0y−gt
  
  y=V0yt−12gt2
  
  Vy^2=V0y^2−2gy
  
  3. The attempt at a solution
  just looking for a quick bit of help here. You can see some of the work I've done in the attached image.
  
  My initial thought was to first decompose the initial velocity vector into its components so I could go ahead and use those in my calculations, which I did, and they seem to check out when compared with the original problems statement on velocity.
  
  Since I'm trying to find the horizontal distance from the roof ledge, I assumed that using this would be the best bet.
  $$(x-x_0)=V_0\cos{\theta}(t)$$
  
  noticing that I need time for this, I decided to use the formula to solve for t
  $$y={V_{0y}}^2t+\frac{at^2}{2}$$
  Where I would let y be -14, since my displacement will be from 14m above 0m (0-14)
  
  however, as you'll see from my work found in the attachment, it looks like I'll keep ending up with complex numbers for time.
  
  I'm wondering whether my assumed initial variables are wrong, or if I've once again made a mathematical blunder?
  
  Any help would be great.
  
  #### Attached Files:
  
  • ###### phy4aH3P3.gif
  File size:
  20 KB
  Views:
  436
  Last edited: Sep 17, 2011
  2. Sep 17, 2011
  
  ### Liquidxlax
  
  Well i miss read the problem at the start and had a few paragraphs describing whats going on, but oh well
  
  The roof is on an incline of 40 degrees and your magnitude of velocity is 7m/s
  
  you need to break this up into it's x and y components using trig
  
  x = vsin(theta)
  
  y = vcos(theta)
  
  Since there are no external forces in the x direction the velocity is constant, but in the y direction there is a gravity.
  
  So by using your kinematic equations and knowing the height of the roof and what gravity is you need to solve for the time it takes for the hammer to reach the ground with your initial y velocity and final y velocity which is zero.
  
  You know g, d, vy1 and vy2 and your x velocity
  
  I missed your picture at first and it looks good to me.
  
  good luck
explanation: The document is a forum post discussing a physics problem related to projectile motion. The user, Nickg140143, presents a problem statement, relevant equations, and their attempt at a solution, asking for help with their calculations. Another user, Liquidxlax, provides feedback and confirms that the approach looks good.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # Projectile Motion - hammer sliding off a roof
  
  1. Sep 17, 2011
  
  ### Nickg140143
  
  1. The problem statement, all variables and given/known data
  A hammer slides off the roof of a house that slopes downward at 40 degrees. The edge of teh roof is 14m avove the ground and the hammer leaves the roof at 7m/s.
  
  How far from the edge of the roof does the hammer strike the ground?
  
  2. Relevant equations
  x direction
  Vx=V0x
  
  x=V0cosθ
  
  y direction
  Vy=V0y−gt
  
  y=V0yt−12gt2
  
  Vy^2=V0y^2−2gy
  
  3. The attempt at a solution
  just looking for a quick bit of help here. You can see some of the work I've done in the attached image.
  
  My initial thought was to first decompose the initial velocity vector into its components so I could go ahead and use those in my calculations, which I did, and they seem to check out when compared with the original problems statement on velocity.
  
  Since I'm trying to find the horizontal distance from the roof ledge, I assumed that using this would be the best bet.
  $$(x-x_0)=V_0\cos{\theta}(t)$$
  
  noticing that I need time for this, I decided to use the formula to solve for t
  $$y={V_{0y}}^2t+\frac{at^2}{2}$$
  Where I would let y be -14, since my displacement will be from 14m above 0m (0-14)
  
  however, as you'll see from my work found in the attachment, it looks like I'll keep ending up with complex numbers for time.
  
  I'm wondering whether my assumed initial variables are wrong, or if I've once again made a mathematical blunder?
  
  Any help would be great.
  
  #### Attached Files:
  
  • ###### phy4aH3P3.gif
  File size:
  20 KB
  Views:
  436
  Last edited: Sep 17, 2011
  2. Sep 17, 2011
  
  ### Liquidxlax
  
  Well i miss read the problem at the start and had a few paragraphs describing whats going on, but oh well
  
  The roof is on an incline of 40 degrees and your magnitude of velocity is 7m/s
  
  you need to break this up into it's x and y components using trig
  
  x = vsin(theta)
  
  y = vcos(theta)
  
  Since there are no external forces in the x direction the velocity is constant, but in the y direction there is a gravity.
  
  So by using your kinematic equations and knowing the height of the roof and what gravity is you need to solve for the time it takes for the hammer to reach the ground with your initial y velocity and final y velocity which is zero.
  
  You know g, d, vy1 and vy2 and your x velocity
  
  I missed your picture at first and it looks good to me.
  
  good luck
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a forum post discussing a physics problem related to projectile motion. The user, Nickg140143, presents a problem statement, relevant equations, and their attempt at a solution, asking for help with their calculations. Another user, Liquidxlax, provides feedback and confirms that the approach looks good.
  
  Label: question-and-answer