Urgent Dynamics Problem Help Needed.

  • Thread starter AEfly
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  • #1
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.
 

Answers and Replies

  • #2
Doc Al
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Hint: How far below A is B?
 
  • #3
AEfly
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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.
 
  • #4
Doc Al
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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.)
 
  • #5
AEfly
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I know VA and t just not how to get VB. Doesnt Vy increase with time in this case as gravity acts downward?
 
  • #6
Doc Al
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Doesnt Vy increase with time in this case as gravity acts downward?
Right. What's the kinematic equation describing that relationship?
 
  • #7
AEfly
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Vy=(Vo)y+at ?
 
  • #9
AEfly
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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
Doc Al
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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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