The End of the Ski Jump - Optimizing Launch Angle

  • #1

Homework Statement


A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/s as shown in Figure 4.14. The landing incline below her falls off with a slope of 35.0°. Where does she land on the incline? I've attached an image of the problem, my work is below it.
RN8sQ5y.png

yzjFy3T.png



Homework Equations


xf = (Vicos(theta))t = dcos(phi)
yf = (Visin(theta)t) - (0.5)gt^2 = -dsin(phi)

The Attempt at a Solution


I understand and reached the solution of part a), but I am struggling with part b). It says to eliminate time "t" and differentiate and maximize "d" in terms of angle "theta".

this is the furthest I've gotten:

xf = (Vicos(theta))t

t = xf/(Vicos(theta))

yf = (Visin(theta))t - (0.5)gt^2

plugging in t = xf/(Vicos(theta))

yf = (Visin(theta))(xf/(Vicos(theta)) - (0.5)g(xf/(Vicos(theta))^2
= xftan(theta) - (0.5)g(xf/(vicos(theta))^2
= xftan(theta) - xf^2(1/(2gVi^2cos^2(theta))

(factoring out and dividing by xf)

yf/xf = tan(theta) - xf/(2gVi^2cos^2(theta))

-dsin(phi)/(dcos(phi)) = tan(theta) - xf/(2gVi^2cos^2(theta))

-tan(phi) = tan(theta) - xf/(2gVi^2cos^2(theta))

I don't know what to do after this step. Any help would be greatly appreciated. Thank you!




 

Answers and Replies

  • #3
What does theta represent in that equation? Think carefully.
Hi Haruspex, I believe it represents the launch angle
 
  • #4
haruspex
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Hi Haruspex, I believe it represents the launch angle
Yes, but angle to what?

Edit: sorry, forget it..... may have misread something. More soon....
 
  • #5
Yes, but angle to what?

Edit: sorry, forget it..... may have misread something. More soon....
The angle in respect to the x-axis? Is it because it's initially at zero? Because φ is 35 degrees ... ?
 
  • #6
Yes, but angle to what?

Edit: sorry, forget it..... may have misread something. More soon....
Oh okay, all good. Thanks for replying to my thread, though!
 
  • #7
haruspex
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= xftan(theta) - (0.5)g(xf/(vicos(theta))^2
= xftan(theta) - xf^2(1/(2gVi^2cos^2(theta))
Check that step. The second line is dimensionally inconsistent.
(This is a useful technique for finding algebraic errors in physics. Check whether the final equation makes sense dimensionally. If it doesn't, do a 'binary chop' to see where it went wrong.)
 
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  • #8
Check that step. The second line is dimensionally inconsistent.
(This is a useful technique for finding algebraic errors in physics. Check whether the final equation makes sense dimensionally. If it doesn't, do a 'binary chop' to see where it went wrong.)
Okay thanks, I'll review my work once I'm home. Also, what do you mean by "binary chop"?
 
  • #9
haruspex
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