[Year 12 Maths] Mechanics - Resisted Motion - I am getting stuck on the last part

[katsicum]

Hi there, I am getting stuck on the last part (part iii) and I don't really understand the question either. When it says "the velocity of the particle", does it mean it's the initial velocity or the escape velocity? I guess I have to use the previous parts and I have tried to used part ii) by letting H approach infinity and hence the denominator be 0 but I couldn't get the answer. Could someone please help me with this part? Thank you so much!
The solution provided by my teacher is 313.7m/s

 

[HallsofIvy]

The problem says "its speed, V, in any position, x" so it clearly does not mean its initial speed or "escape velocity (which would be the initial speed necessary to "escape" the earth's gravitational pull)! You are told that its initial speed is "u" and the third part of the problem asks you to find "escape velocity".

We are given that \(\displaystyle \frac{d}{dx}\left(\frac{1}{2}v^2\right)= -\frac{k}{x^2}\). So \(\displaystyle d(\frac{1}{2}v^2)= k\frac{dx}{x^2}\). Integrate both sides. There will, of course, be an unknown "constant of integration". To determine it use the fact that when x= R, v= u.

As long as the object is going up, it speed will be positive. When it is coming down, it speed is negative. At the very top, it stops going up but is not yet coming down so its speed is 0. Solve \(\displaystyle u^2- 2gR^2\left(\frac{1}{R}- \frac{1}{x}\right)= 0\) for x.

Yes, to find "escape velocity" you need to determine "u" so that H is arbitrarily large and you can do that by taking H to be infinite. For a fraction, \(\displaystyle \frac{p}{q}\), to be "infinite" the denominator, q, must be 0.