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

Commented by ajfour last updated on 01/Jan/21

A ball is dropped from a tall tower, height h, at the equator. Find how far from the foot of the building, does the ball land.

$${A}\:{ball}\:{is}\:{dropped}\:{from}\:{a}\:{tall} \\ $$$${tower},\:{height}\:{h},\:{at}\:{the}\:{equator}. \\ $$$${Find}\:{how}\:{far}\:{from}\:{the}\:{foot}\:{of} \\ $$$${the}\:{building},\:{does}\:{the}\:{ball}\:{land}. \\ $$

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

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

the ball moves in horizontal direction with the same velocity as the top of the tower which is larger than the speed of the foot of the tower. BC=AA′>BB′. that means the balls falls on ground in front of the foot of the tower, not behind it.

$${the}\:{ball}\:{moves}\:{in}\:{horizontal}\:{direction} \\ $$$${with}\:{the}\:{same}\:{velocity}\:{as}\:{the}\:{top} \\ $$$${of}\:{the}\:{tower}\:{which}\:{is}\:{larger}\:{than} \\ $$$${the}\:{speed}\:{of}\:{the}\:{foot}\:{of}\:{the}\:{tower}. \\ $$$${BC}={AA}'>{BB}'. \\ $$$${that}\:{means}\:{the}\:{balls}\:{falls}\:{on}\:{ground} \\ $$$${in}\:{front}\:{of}\:{the}\:{foot}\:{of}\:{the}\:{tower}, \\ $$$${not}\:{behind}\:{it}. \\ $$

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

u=(R+h)ω h=(1/2)gt^2 t=(√((2h)/g)) BC=AA′=ut=(R+h)ω(√((2h)/g)) BB′=Rω(√((2h)/g)) s=BC−BB′=hω(√((2h)/g)) example: ω=((2π)/(24×60×60)) h=100 m s=100×((2π)/(24×60×60))×(√((2×100)/(10))) ≈0.0325 m=32.5 mm

$${u}=\left({R}+{h}\right)\omega \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${t}=\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$${BC}={AA}'={ut}=\left({R}+{h}\right)\omega\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$${BB}'={R}\omega\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$${s}={BC}−{BB}'={h}\omega\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$${example}: \\ $$$$\omega=\frac{\mathrm{2}\pi}{\mathrm{24}×\mathrm{60}×\mathrm{60}} \\ $$$${h}=\mathrm{100}\:{m} \\ $$$${s}=\mathrm{100}×\frac{\mathrm{2}\pi}{\mathrm{24}×\mathrm{60}×\mathrm{60}}×\sqrt{\frac{\mathrm{2}×\mathrm{100}}{\mathrm{10}}} \\ $$$$\approx\mathrm{0}.\mathrm{0325}\:{m}=\mathrm{32}.\mathrm{5}\:{mm} \\ $$

Commented by ajfour last updated on 01/Jan/21

thanks Sir, I too got an answer close to yours!

$${thanks}\:{Sir},\:{I}\:{too}\:{got}\:{an}\:{answer} \\ $$$${close}\:{to}\:{yours}! \\ $$

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

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

in my consideration above the influence of the negative acceleration in horizontal direction g_x is not taken into account (the dotted line in following diagram). when this is considered we′ll get the correct formula s=(2/3)hω(√((2h)/g))

$${in}\:{my}\:{consideration}\:{above}\:{the}\: \\ $$$${influence}\:{of}\:{the}\:{negative}\:{acceleration} \\ $$$${in}\:{horizontal}\:{direction}\:{g}_{{x}} \:{is}\:{not} \\ $$$${taken}\:{into}\:{account}\:\left({the}\:{dotted}\:{line}\right. \\ $$$$\left.{in}\:{following}\:{diagram}\right).\:{when}\:{this}\:{is} \\ $$$${considered}\:{we}'{ll}\:{get}\:{the}\:{correct} \\ $$$${formula} \\ $$$${s}=\frac{\mathrm{2}}{\mathrm{3}}{h}\omega\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$

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=((u(R+h))/r) (dx/dt)=((u(R+h))/((R+h)−((gt^2 )/2))) ∫dx=−((2u(R+h))/g)∫_0 ^( T) (dt/(t^2 −((2(R+h))/g))) s=((2u(R+h))/g)×((√g)/(2(√(2(R+h)))))ln ∣(((√((2(R+h))/g))+t)/( (√((2(R+h))/g))−t))∣_0 ^T s=u(√((R+h)/(2g)))ln (1+((2T)/λ)) ≈u(√((R+h)/(2g)))(2(√((2h)/g)))(√(g/(2(R+h)))) ≈ ω(R+h)(√((2h)/g)) =(((2π)/(1day)))(R+h)(√((2h)/g)) s_(tower) =(√((2h)/g))(((2π)/(1day)))R △=(((2π)/(86400)))h(√((2h)/g)) for h=100m △≈((π((√5)))/(216))×100cm =3.25cm .

$${mVr}={mu}\left({R}+{h}\right) \\ $$$${V}=\frac{{u}\left({R}+{h}\right)}{{r}} \\ $$$$\frac{{dx}}{{dt}}=\frac{{u}\left({R}+{h}\right)}{\left({R}+{h}\right)−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\int{dx}=−\frac{\mathrm{2}{u}\left({R}+{h}\right)}{{g}}\int_{\mathrm{0}} ^{\:{T}} \frac{{dt}}{{t}^{\mathrm{2}} −\frac{\mathrm{2}\left({R}+{h}\right)}{{g}}} \\ $$$${s}=\frac{\mathrm{2}{u}\left({R}+{h}\right)}{{g}}×\frac{\sqrt{{g}}}{\mathrm{2}\sqrt{\mathrm{2}\left({R}+{h}\right)}}\mathrm{ln}\:\mid\frac{\sqrt{\frac{\mathrm{2}\left({R}+{h}\right)}{{g}}}+{t}}{\:\sqrt{\frac{\mathrm{2}\left({R}+{h}\right)}{{g}}}−{t}}\mid_{\mathrm{0}} ^{{T}} \\ $$$${s}={u}\sqrt{\frac{{R}+{h}}{\mathrm{2}{g}}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{2}{T}}{\lambda}\right) \\ $$$$\:\:\approx{u}\sqrt{\frac{{R}+{h}}{\mathrm{2}{g}}}\left(\mathrm{2}\sqrt{\frac{\mathrm{2}{h}}{{g}}}\right)\sqrt{\frac{{g}}{\mathrm{2}\left({R}+{h}\right)}} \\ $$$$\:\:\approx\:\omega\left({R}+{h}\right)\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$$\:\:=\left(\frac{\mathrm{2}\pi}{\mathrm{1}{day}}\right)\left({R}+{h}\right)\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$$\:\:{s}_{{tower}} =\sqrt{\frac{\mathrm{2}{h}}{{g}}}\left(\frac{\mathrm{2}\pi}{\mathrm{1}{day}}\right){R} \\ $$$$\bigtriangleup=\left(\frac{\mathrm{2}\pi}{\mathrm{86400}}\right){h}\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$$\:{for}\:\:{h}=\mathrm{100}{m} \\ $$$$\bigtriangleup\approx\frac{\pi\left(\sqrt{\mathrm{5}}\right)}{\mathrm{216}}×\mathrm{100}{cm}\:=\mathrm{3}.\mathrm{25}{cm}\:. \\ $$

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

your result Δ=ωh(√((2h)/g)) is the same as what i did at the begining. but it is not correct, because the acceleration g is always toward the center of the earth and has therefore a component in x−direction. this component g_x reduces the horizontal velocity and therefore the horizontal displacement. the correct answer is Δ=(2/3)ωh(√((2h)/g)).

$${your}\:{result}\:\Delta=\omega{h}\sqrt{\frac{\mathrm{2}{h}}{{g}}}\:{is}\:{the}\:{same} \\ $$$${as}\:{what}\:{i}\:{did}\:{at}\:{the}\:{begining}.\:{but}\:{it} \\ $$$${is}\:{not}\:{correct},\:{because}\:{the}\:{acceleration} \\ $$$${g}\:{is}\:{always}\:{toward}\:{the}\:{center}\:{of}\:{the} \\ $$$${earth}\:{and}\:{has}\:{therefore}\:{a}\:{component} \\ $$$${in}\:{x}−{direction}.\:{this}\:{component}\:{g}_{{x}} \\ $$$${reduces}\:{the}\:{horizontal}\:{velocity}\:{and} \\ $$$${therefore}\:{the}\:{horizontal}\: \\ $$$${displacement}.\:{the}\:{correct}\:{answer}\:{is} \\ $$$$\Delta=\frac{\mathrm{2}}{\mathrm{3}}\omega{h}\sqrt{\frac{\mathrm{2}{h}}{{g}}}. \\ $$

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