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

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The discussion focuses on solving a dynamics problem from Hibbler's 13th edition, specifically determining the initial speed (VA) of a skier leaving a ramp at a 15-degree angle and the speed (VB) at which he strikes the ground. The user understands how to calculate VA but struggles with finding VB, needing guidance on setting up the equations correctly. It is suggested to use the time derived from the first equation to find the vertical and horizontal components of VB. The relationship between vertical velocity (Vy) and time is emphasized, with the kinematic equation provided as a key tool for solving the problem. Ultimately, the solution involves calculating both components of VB and combining them to find the resultant speed.
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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