Projectile problem-A rock is kicked off a 45 Degree hill at 15m/s.

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Homework Help Overview

The problem involves a rock being kicked horizontally off a 45-degree hill at a speed of 15 m/s. The objective is to determine the time it takes for the rock to hit the ground, considering its initial horizontal velocity and the slope of the hill.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the x and y position functions and their intersection, questioning how to account for the hill's slope in the calculations. There is also mention of the initial conditions and how they affect the trajectory of the rock.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup. Some guidance has been offered regarding the relationship between the x and y positions, but there is no explicit consensus on the correct approach or solution.

Contextual Notes

There is mention of a potential misprint in the problem statement regarding the initial velocity, which could affect the answers being derived. Participants are also considering the implications of launching the rock horizontally versus off the slope itself.

Kizaru
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Projectile problem--A rock is kicked off a 45 Degree hill at 15m/s.

Hello. I tried this problem and it's the last one for my problem set due Thursday, but my answer differs from the given answer in the back of the book. After trying the solution again, I searched Google and found this. I'm having the exact problem.

Homework Statement


A rock is kicked horizontally at 15m/s from a hill with a 45 degree slope. How long does it take for the rock to hit the ground?
Vix = 15m/s
Viy = 0
Yi = 0
Xi = 0

Homework Equations


Vx = Vix
X = Xi + Vxo*t
Vy = Viy - gt
Y = Yi + Viy*t - (1/2)g*t^2


The Attempt at a Solution


I tried equating the X and Y position functions, but this only provides the intersection between them--it does not provide the intersection between the rock and the hill. My second method is identical to the one described in the link mentioned earlier in my post. The problem is that my answer isn't the same as the one in the back of the book, and I'm not sure if it's a problem on my end. I don't know what to do after this. Any help would be great :)
 
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If the rock lands at the same height it was launched one can obtain the time from the y position equation since y=0.
 


Kurdt said:
If the rock lands at the same height it was launched one can obtain the time from the y position equation since y=0.

The rock will never pass that point because it's launched horizontally (no initial Y velocity) off a 45 degree hill (the hill is declining). The only time the rock is at y=0 is when t=0.

Imagine graphing y=-x and y=-x^2. If you focus on only the third quadrant, you have what looks like the hill (y=-x) and the path of the rock (y=-x^2), but it's not that easy to find an equation that represents the path of the rock.
 


Oh ok I get it. I thought it was launched off a 45 degree slope.

Well you know how the x position and y position are related (equation of slope) and you know how they evolve over time (kinematic equations). You can use both these bits of info to find the time.
 


For those getting an answer of 3.06 seconds, which disagrees with the back of the book, this is because there was a misprint. The actual velocity should be 25 m/s, not 15 m/s. The solutions manual also uses 25 m/s. Good luck!
 

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