The End of the Ski Jump - Optimizing Launch Angle

[Bdhillon1994]

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.

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!

 

[haruspex]

xf = (Vicos(theta))t

What does theta represent in that equation? Think carefully.

 

[Bdhillon1994]

What does theta represent in that equation? Think carefully.

Hi Haruspex, I believe it represents the launch angle

 

[haruspex]

Hi Haruspex, I believe it represents the launch angle

Yes, but angle to what?

Edit: sorry, forget it... may have misread something. More soon...

 

[Bdhillon1994]

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 ... ?

 

[Bdhillon1994]

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!

 

[haruspex]

= 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.)

 

[Bdhillon1994]

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"?

 

[haruspex]

Okay thanks, I'll review my work once I'm home. Also, what do you mean by "binary chop"?

Say you did eight operations on an equation in sequence. The first equation passes some check (such as dimensional consistency) but the ninth doesn't. so check the one in the middle. If that's ok, check the one half way from there to the end, etc.
You might find this useful: https://www.physicsforums.com/insights/frequently-made-errors-equation-handling

 

[Bdhillon1994]

Say you did eight operations on an equation in sequence. The first equation passes some check (such as dimensional consistency) but the ninth doesn't. so check the one in the middle. If that's ok, check the one half way from there to the end, etc.
You might find this useful: https://www.physicsforums.com/insights/frequently-made-errors-equation-handling

Thanks!