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Question Number 127708 by ajfour last updated on 01/Jan/21

Commented by ajfour last updated on 01/Jan/21

$$Aballisdroppedfromatall$$$$tower,heighth,attheequator.$$$$Findhowfarfromthefootof$$$$thebuilding,doestheballland.$$

Commented by mr W last updated on 01/Jan/21

Commented by mr W last updated on 01/Jan/21

$$theballmovesinhorizontaldirection$$$$withthesamevelocityasthetop$$$$ofthetowerwhichislargerthan$$$$thespeedofthefootofthetower.$$$$BC={AA}^{\prime }>{BB}^{\prime }.$$$$thatmeanstheballsfallsonground$$$$infrontofthefootofthetower,$$$$notbehindit.$$

Commented by mr W last updated on 02/Jan/21

$$u=(R+h)\omega$$$$h=\frac{1}{2}{gt}^2$$$$t=\sqrt{\frac{2h}{g}}$$$$BC={AA}^{\prime }=ut=(R+h)\omega \sqrt{\frac{2h}{g}}$$$${BB}^{\prime }=R\omega \sqrt{\frac{2h}{g}}$$$$s=BC-{BB}^{\prime }=h\omega \sqrt{\frac{2h}{g}}$$$$example:$$$$\omega =\frac{2\pi }{24\times 60\times 60}$$$$h=100m$$$$s=100\times \frac{2\pi }{24\times 60\times 60}\times \sqrt{\frac{2\times 100}{10}}$$$$\approx 0.0325m=32.5mm$$

Commented by ajfour last updated on 01/Jan/21

$$thanksSir,Itoogotananswer$$$$closetoyours!$$

Commented by mr W last updated on 01/Jan/21

Commented by mr W last updated on 01/Jan/21

$$inmyconsiderationabovethe$$$$influenceofthenegativeacceleration$$$$inhorizontaldirection{g}_{x}isnot$$$$takenintoaccount(thedottedline$$$$infollowingdiagram).whenthisis$$$$considered{we}^{\prime }llgetthecorrect$$$$formula$$$$s=\frac{2}{3}h\omega \sqrt{\frac{2h}{g}}$$

Commented by mr W last updated on 01/Jan/21

Commented by ajfour last updated on 01/Jan/21

Commented by ajfour last updated on 01/Jan/21

$$mVr=mu(R+h)$$$$V=\frac{u(R+h)}{r}$$$$\frac{dx}{dt}=\frac{u(R+h)}{(R+h)-\frac{{gt}^2}{2}}$$$$\int dx=-\frac{2u(R+h)}{g}{\int }_0^T\frac{dt}{t^2-\frac{2(R+h)}{g}}$$$$s=\frac{2u(R+h)}{g}\times \frac{\sqrt{g}}{2\sqrt{2(R+h)}}\ln \mid \frac{\sqrt{\frac{2(R+h)}{g}}+t}{\sqrt{\frac{2(R+h)}{g}}-t}{\mid }_0^T$$$$s=u\sqrt{\frac{R+h}{2g}}\ln (1+\frac{2T}{\lambda })$$$$\approx u\sqrt{\frac{R+h}{2g}}\left(2\sqrt{\frac{2h}{g}}\right)\sqrt{\frac{g}{2(R+h)}}$$$$\approx \omega (R+h)\sqrt{\frac{2h}{g}}$$$$=\left(\frac{2\pi }{1day}\right)(R+h)\sqrt{\frac{2h}{g}}$$$$s_{tower}=\sqrt{\frac{2h}{g}}\left(\frac{2\pi }{1day}\right)R$$$$△=\left(\frac{2\pi }{86400}\right)h\sqrt{\frac{2h}{g}}$$$$forh=100m$$$$△\approx \frac{\pi \left(\sqrt{5}\right)}{216}\times 100cm=3.25cm.$$

Commented by mr W last updated on 01/Jan/21

$$yourresult\mathrm{\Delta }=\omega h\sqrt{\frac{2h}{g}}isthesame$$$$aswhatididatthebegining.butit$$$$isnotcorrect,becausetheacceleration$$$$gisalwaystowardthecenterofthe$$$$earthandhasthereforeacomponent$$$$inx-direction.thiscomponent{g}_x$$$$reducesthehorizontalvelocityand$$$$thereforethehorizontal$$$$displacement.thecorrectansweris$$$$\mathrm{\Delta }=\frac{2}{3}\omega h\sqrt{\frac{2h}{g}}.$$

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