I am in full agreement with Newzad. Here's my solution -- Given: y=x^2/4-2x+8, dy/dx = x/2 - 2 which for maximum range must equal tan(α) = tan(45) = 1. This gives us: x/2 -2 = 1 or x = 6 or B is (6,5). Now with B as origin we've: y = xtan(α)-(g/2)(x/ucos(α))^2 where α = 45 deg. and u^2= 2*g*(8 - 5)= 6*g or y= x - x^2/6. In the original frame of reference, our equation: y - 5 =(x -6)-(x -6)^2/6 If we set y=0 in this equation of parabolic motion, we get x= (9 - V39) or (9 + V39) where V = square root. The first root is clearly inadmissible hence x = 9 + V39 ~ 15.245 Ajit

The options will be updated. Thanks for your input. I made a simple error: Instead of R = x + Vx(t1+t2), I wrote R = x + Vx(t2), that's why 12 or 13 is very short.

BUT the solution is NOT 15.245. Believe me the solution is NOT 15.245. Still not confirmed yet but I have a solution bigger than 15.245. It's amazing! Good luck !

Hi, Along the lines suggested by you, I set up the following equation:-y=z(R-x)-u(R-x)^2/(4(8-y)) where y=x^2/4-2x +8, tan(α)=z=x/2-2 & (sec(α))^2=u=1+z^2 and R is the range. By varying x, I found that at x=16/3, y=40/9 the range is exactly 16 and I think that's the maximum. In our equation if we put x=6 we get R=15.2445 as b4 which is clearly not the highest possible. Thanks for the nice problem! Ajit

Yes Yuv-k, 16 is the answer. But it doesn't depend on g. As long as there is a positive gravity, the answer is the same. In other words, Max range is 16 on Earth, Jupiter as well as on Pluto.

As a side note: those who degraded Pluto from planet to dwarf will pay dearly for their arrogance :(

I am in full agreement with Newzad. Here's my solution -- Given: y=x^2/4-2x+8, dy/dx = x/2 - 2 which for maximum range must equal tan(α) = tan(45) = 1. This gives us: x/2 -2 = 1 or x = 6 or B is (6,5). Now with B as origin we've: y = xtan(α)-(g/2)(x/ucos(α))^2 where α = 45 deg. and u^2= 2*g*(8 - 5)= 6*g or y= x - x^2/6. In the original frame of reference, our equation:

ReplyDeletey - 5 =(x -6)-(x -6)^2/6

If we set y=0 in this equation of parabolic motion, we get x= (9 - V39) or (9 + V39) where V = square root. The first root is clearly inadmissible hence x = 9 + V39 ~ 15.245

Ajit

This comment has been removed by the author.

ReplyDeleteThe options will be updated. Thanks for your input. I made a simple error: Instead of R = x + Vx(t1+t2), I wrote R = x + Vx(t2), that's why 12 or 13 is very short.

ReplyDeleteBUT the solution is NOT 15.245.

Believe me the solution is NOT 15.245.

Still not confirmed yet but I have a solution bigger than 15.245. It's amazing!

Good luck !

Hi,

ReplyDeleteAlong the lines suggested by you, I set up the following equation:-y=z(R-x)-u(R-x)^2/(4(8-y)) where y=x^2/4-2x +8, tan(α)=z=x/2-2 & (sec(α))^2=u=1+z^2 and R is the range. By varying x, I found that at x=16/3, y=40/9 the range is exactly 16 and I think that's the maximum. In our equation if we put x=6 we get R=15.2445 as b4 which is clearly not the highest possible.

Thanks for the nice problem!

Ajit

YES, 16 is the answer for the maximum range.

ReplyDeleteThank you Ajit.

Yes 16 is answer (for g=10).

ReplyDeleteYes Yuv-k, 16 is the answer. But it doesn't depend on g. As long as there is a positive gravity, the answer is the same. In other words, Max range is 16 on Earth, Jupiter as well as on Pluto.

ReplyDeleteAs a side note: those who degraded Pluto from planet to dwarf will pay dearly for their arrogance :(