brotherbobby's latest activity

Yes I understand. I must admit that I have often had the urge to come on here for problems which proved difficult after a few attempts...

Yes, I suppose it's a mistake to come on here too soon.

Thanks. Corrected.

Aha I remember ##(x+y)^2-(x-y)^2=4xy\Rightarrow xy = \dfrac{1}{4}\{(x+y)^2-(x-y)^2\}##. Worth trying (2nd method above)...

Let me start by showing the problem as it appeared in the text. Attempt : ##{\color{red}{\large *}}## The key thing is to do the...

Ok so the horizontal distance from the point of projection to the point of landing is 20m?

Yes. Which means if A cycles a distance ##d## greater than he walks, B has to walk the distance ##d## greater than he gets to cycle and...

In your problem statement, you wrote "The flight distance D = 20m". What distance is this? The horizontal distance or vertical? Or is...

Yes I had already written out the solution in post#14, even though I had conceptual doubts.

Brilliant. I was about to do it myself but thanks a lot. It was a tricky problem and I have clarified my doubts.

I have drawn the distance time graph of the problem below for tourists A and B, and the cycle C. But it hasn't helped me in my solution...

Thank you all for your comments and apologies for the almost 24 hour delay in my reply. For now @haruspex 's method seems the most...

By simplest do you mean the number of turns they take on the bicycle? In that case I can think of a single turn walking and bicycle for...

Yes, I agree. Since the bicycle is left idle, walking times increase over cycle times making ##t_W>t_C##. I cannot think of a pattern...

The simplest pattern would be that they travel either half the distance, or half the time, walking and cycling. Since...