Minimum speed to cross ravine

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  • #1

Homework Statement


A motorcyclist must cross a ravine. The far side of the ravine is 2.0 meters higher than the launch point, a ramp that makes an angle of 40◦ with the horizontal. If the ravine is 10.0 meters wide, what minimum speed v must the biker have when leaving the ramp to successfully cross the ravine? Take the launch point at the end of the ramp as the coordinate origin.

Homework Equations


yf=yi+(vi)t-.5g(t^2)

The Attempt at a Solution



Well, I know that Vxi=vicos40 and vyi=visin40.
Known: xi=0, yi=0, ti=0, xf=10, yf=2, theta=40, ay=-9.8
I want to use to equation above, but I can't because I know neither the Vi nor the T. I thought I could figure out the T by saying that the initial vy was zero, but that definitely doesn't seem right because then only gravity would be acting on the bike and he has to GAIN 2m in height. So I'm a little stuck. I'd really appreciate any tips to help me figure out this problem!
 

Answers and Replies

  • #2
haruspex
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You have two equations, one in the y direction and one in the x direction. There are two unknowns they share.
 
  • #3
The only equation in the x direction I can think of is xf=xi+(vx)i*t, and here I still have two unknowns (xfi and t).
 
  • #4
haruspex
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The only equation in the x direction I can think of is xf=xi+(vx)i*t, and here I still have two unknowns (xfi and t).
No, the two unknowns there are vx and t, surely? And you can express vx in terms of v and theta. So as I posted, both unknowns appear in both equations. Two equations, two unknowns.
 
  • #5
Ah, yes, I definitely meant Vx and t for the unknowns.

So, if I say that vx=vicos40 then I can rewrite the equation as xf=xi+(vicos40)t... which can be rearranged to say t=xf]-xi/(vicos40).

then I think I can substitute into the first equation:
yf=yi+(visin40)(xf-xi/vicos40)+.5(-9.8)(xf-xi/vicos40)^2. So my only unkown is vi! ...Now, if only I can get the math right haha..

Ok, I calculated that Vi=9.92m/s. I think this makes sense. Thanks for helping me, I never would have thought to substitute the equations on my own.
 
  • #6
PeroK
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Ah, yes, I definitely meant Vx and t for the unknowns.

So, if I say that vx=vicos40 then I can rewrite the equation as xf=xi+(vicos40)t... which can be rearranged to say t=xf]-xi/(vicos40).

then I think I can substitute into the first equation:
yf=yi+(visin40)(xf-xi/vicos40)+.5(-9.8)(xf-xi/vicos40)^2. So my only unkown is vi! ...Now, if only I can get the math right haha..

Ok, I calculated that Vi=9.92m/s. I think this makes sense. Thanks for helping me, I never would have thought to substitute the equations on my own.

9.92m/s isn't fast enough. You'll end up in the ravine at that speed! Maybe check the maths.
 
  • #7
haruspex
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Ah, yes, I definitely meant Vx and t for the unknowns.

So, if I say that vx=vicos40 then I can rewrite the equation as xf=xi+(vicos40)t... which can be rearranged to say t=xf]-xi/(vicos40).

then I think I can substitute into the first equation:
yf=yi+(visin40)(xf-xi/vicos40)+.5(-9.8)(xf-xi/vicos40)^2. So my only unkown is vi! ...Now, if only I can get the math right haha..

Ok, I calculated that Vi=9.92m/s. I think this makes sense. Thanks for helping me, I never would have thought to substitute the equations on my own.
I haven't checked your arithmetic, but good to see that you get the method.
 

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