Urgent Dynamics Problem Help Needed: 12-101 Hibbler 13th ed.

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Discussion Overview

The discussion revolves around a dynamics problem from the 13th edition of Hibbler, specifically focusing on a skier leaving a ramp at an angle and determining both the initial speed and the speed upon striking the ground. The scope includes mathematical reasoning and problem-solving related to kinematics and projectile motion.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant outlines the problem and expresses uncertainty about calculating the speed at which the skier strikes the ground (VB).
  • Another participant asks for the vertical distance between points A and B, which is confirmed to be 64m.
  • A participant shares their understanding of the equations for VA and expresses confusion about how to set up the equations for VB.
  • There is a discussion about the relationship between vertical velocity (Vy) and time, with participants confirming that Vy increases due to gravity.
  • One participant proposes using the kinematic equation to relate Vy to time and initial velocity.
  • Another participant suggests that to find VB, both vertical (VBy) and horizontal (VBx) components need to be calculated, followed by using the Pythagorean theorem to find the resultant speed.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving for VB, but there is no consensus on the specific steps or equations to use, indicating that the discussion remains unresolved.

Contextual Notes

Participants express uncertainty about the setup of equations and the relationships between variables, indicating potential limitations in their understanding of the problem's kinematics.

AEfly
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The Problem.
12-101 Hibbler dynamics 13th edition.

It is observed that the skier leaves the ramp A at an angle θA=15 degrees with the horizontal. if he strikes the ground at B, deterime his initial speed VA and the speed at which he strikes the ground (at B).

http://s3.amazonaws.com/answer-board-image/d324abb1-8b50-49a3-a0dd-e2375f1be1a4.jpeg (Ignore the values not in the diagram itself)

My issue:
I understand how to get VA but, I have no idea how to the the speed at which he strikes the ground at B. Any help would be very appreciated.

I do not even know where to start to find VB.
 
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Hint: How far below A is B?
 
64m. I just don't know how to set up and re work the equations to solve for VB. I understand for VA its

80=VA(cos15)t

-64=VA(sin15)t-(1/2)(9.81)t^2

where I would sub in VA=80/cos15(t) to solve for t and then plug t back into the first equation.
 
AEfly said:
64m. I just don't know how to set up and re work the equations to solve for VB. I understand for VA its

80=VA(cos15)t

-64=VA(sin15)t-(1/2)(9.81)t^2

where I would sub in VA=80/cos15(t) to solve for t and then plug t back into the first equation.
Since you know VA, you can use either equation to solve for t. (Use the first one, of course.)

How does Vy depend on time?

(There are several ways to get the answer; one way just uses the distance, but since you have the time why not use it.)
 
I know VA and t just not how to get VB. Doesnt Vy increase with time in this case as gravity acts downward?
 
AEfly said:
Doesnt Vy increase with time in this case as gravity acts downward?
Right. What's the kinematic equation describing that relationship?
 
Vy=(Vo)y+at ?
 
AEfly said:
Vy=(Vo)y+at ?
Sure.
 
But how does VB relate into that? I am guessing I need to find VBy and VBx and then do the

sqrt of (VBy)^2 + (VBx)^2
 
  • #10
AEfly said:
But how does VB relate into that? I am guessing I need to find VBy and VBx and then do the

sqrt of (VBy)^2 + (VBx)^2
Exactly. VBx should be trivial.
 

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