Hi there, I am so close to getting over this question but unfortunately I am still getting stuck on the last part. I have tried to a lot of ways but none has worked out. Attached below are the question, the answer to part iv) along with my full working out for the previous parts respectively. Thank you so so much for helping me!
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Mechanics - Resisted Motion - Getting stuck on the last part (part iv) -Please help! (1 Viewer)
- Thread starter catha230
- Start date
You have already established, for the upwards motion, that:
:} \qquad \ddot{x} = \cfrac{-g\big(v^2 + V^2\big)}{V^2})
:} \qquad \ln \Bigg| \cfrac{v^2 + V^2}{V^2} \Bigg| = \cfrac{g}{V^2} \big(H - x\big))
:} \qquad H = \cfrac{V^2 \ln 2}{2g})
:} \qquad v^2 = V^2 \Big(2e^{\frac{-2gx}{V^2}} - 1\Big) = V^2 \Bigg(2\exp\bigg(\cfrac{-2gx}{V^2}\bigg) - 1\Bigg))
:} \qquad t= \cfrac{V}{g}\bigg(\cfrac{\pi}{4} - \tan^{-1}\cfrac{v}{V} \bigg))
:} \qquad \cfrac{2V^2}{v^2 + V^2} = 1 + \sin {\cfrac{2gt}{V})
To do part (iv), you need to:
To do part (iv), you need to:
- Use (4A) (or more easily, (4)) and the fact that when
,
to establish that
- Establish that the velocity at half the maximum height is given by by putting
into (2) or, alternatively, by using (2A) with
- Use this velocity in (4A) to establish that the time when the height is half of its maximum is given by
- The required percentage can then be determined by looking at the ratio of
to
, either in exact form or as a decimal approximation:
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