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Angle and range of racer

  • Thread starter dranger35
  • Start date
  • #1
14
0
Need Help 2 problems!!!!

Homework Problem 1:
You are starting out a race to go from point A to point B, and then
back. You have to stop on your return to point A. The initial mass
of your rocket is 80% fuel. Racer 1, who is not as smart as you
and is not taking this class, plans to use ¼ of the fuel to speed
up, ¼ of the fuel to come to a stop at point B, ¼ of the fuel to
speed back up, and the final ¼ of the fuel to stop at point A.
● A) You have a better strategy. What is it?
● B) Assuming the distance between A and B is long - you coast
most of the way - how much faster are you compared to racer
1?
● C) Does Racer 1 have enough fuel left to come to a stop at the
finish point A? If not, what is Racer 1's final speed (v/u). If so,
how much fuel is left over?

Homework Problem 2:
Assume an initial speed of 100 m/s, initial height of 100 m, and
work out the angle that gives the longest trajectory. Give the
angle as xx.xx degrees and the range (horizontal distance
traveled) as yyyy m. Do this numerically, on computer.
 

Answers and Replies

  • #2
298
0
Problem 1
Well, im not entirely sure with #1, but heres a conjecture. There are four basic intervals, each with 25% of the fuel being burned for Racer 1. So, in the first interval, there will be some mass loss and velocity increase. Now, you'll notice that, the force will increase exponentially, as the mass goes down, the acceleration will increase. So, in the first interval, he wastes 25% fuel speeding up. In the second interval, he wastes another 25% to slow down. However, the force is actually much more in this interval than it is in the first interval. This is because the mass has decreased considerably. The more energy efficient way to slow down would be to use a smaller portion of the fuel, which will stop you just as well, but will leave more fuel for the return journey. The same idea occurs on the way back. So i guess the idea would be, you need less fuel to stop than you do to speed up because there is less mass. Not sure if thats right, but it gives you something to think about.

Problem 2
The easiest way i can see to do this on a computer would be to have a huge loop. It starts at an angle 00.01 and goes up to 90.00. You can constantly recalculate the trajectory and then if its longer than any previous value just record it. Just keep re-recording the highest value. It seems very inefficient and im sure there is a more efficient way to do it but thats an approach.
 
  • #3
14
0
Thanks

Hey thanks alot for giving me some insight into the problem. I wasnt sure where to start.
 

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