Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 169969 by Beginner last updated on 13/May/22

Answered by mr W last updated on 13/May/22

Commented by ajfour last updated on 13/May/22

u=(√(2gh)) (usin θ)t+(((gsin θ)t^2 )/2)=s (ucos θ)t−(((gcos θ)t^2 )/2)=0 ⇒ t=((2u)/g) s=((2u^2 sin θ)/g)+((2u^2 sin θ)/g) s=((4(2gh)sin θ)/g) s=8hsin θ

$${u}=\sqrt{\mathrm{2}{gh}} \\ $$$$\left({u}\mathrm{sin}\:\theta\right){t}+\frac{\left({g}\mathrm{sin}\:\theta\right){t}^{\mathrm{2}} }{\mathrm{2}}={s} \\ $$$$\left({u}\mathrm{cos}\:\theta\right){t}−\frac{\left({g}\mathrm{cos}\:\theta\right){t}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}=\frac{\mathrm{2}{u}}{{g}} \\ $$$$\:{s}=\frac{\mathrm{2}{u}^{\mathrm{2}} \mathrm{sin}\:\theta}{{g}}+\frac{\mathrm{2}{u}^{\mathrm{2}} \mathrm{sin}\:\theta}{{g}} \\ $$$$\:{s}=\frac{\mathrm{4}\left(\mathrm{2}{gh}\right)\mathrm{sin}\:\theta}{{g}} \\ $$$$\:{s}=\mathrm{8}{h}\mathrm{sin}\:\theta \\ $$

Commented by mr W last updated on 13/May/22

great sir!

$${great}\:{sir}! \\ $$

Commented by learner22 last updated on 13/May/22

bravo

$$\boldsymbol{{bravo}} \\ $$

Commented by mr W last updated on 13/May/22

Commented by mr W last updated on 13/May/22

assume that the ball hits the plane again, then with λ=(a/h)=((l sin θ)/h) and ξ=(l/h) we have cos 2θ+(√(cos^2 2θ+ξ sin θ))=((ξ sin θ)/(2(1−cos 2θ))) (√(cos^2 2θ+ξ sin θ))=((ξ sin θ)/(2(1−cos 2θ)))−cos 2θ ⇒ξ=8 sin θ

$${assume}\:{that}\:{the}\:{ball}\:{hits}\:{the}\:{plane} \\ $$$${again},\:{then} \\ $$$${with}\:\lambda=\frac{{a}}{{h}}=\frac{{l}\:\mathrm{sin}\:\theta}{{h}}\:{and}\:\xi=\frac{{l}}{{h}}\:{we}\:{have} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta+\sqrt{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta+\xi\:\mathrm{sin}\:\theta}=\frac{\xi\:\mathrm{sin}\:\theta}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)} \\ $$$$\sqrt{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta+\xi\:\mathrm{sin}\:\theta}=\frac{\xi\:\mathrm{sin}\:\theta}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)}−\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Rightarrow\xi=\mathrm{8}\:\mathrm{sin}\:\theta \\ $$

Commented by mr W last updated on 13/May/22

say impact point B is a above the ground. speed at point B: v=(√(2gh)) upon point B: x=v sin 2θ t y=a+v cos 2θ t−((gt^2 )/2) y=a+v cos 2θ×(x/(v sin 2θ))−(g/2)×((x/(v sin 2θ)))^2 y=a+(x/(tan 2θ))−((gx^2 )/(2v^2 sin^2 2θ)) ⇒y=a+(x/(tan 2θ))−(x^2 /(4h sin^2 2θ)) at y=0: a+(x/(tan 2θ))−(x^2 /(4h sin^2 2θ))=0 x^2 −((4h sin^2 2θx)/(tan 2θ))−4ah sin^2 2θ=0 x=((2h sin^2 2θ)/(tan 2θ))+2 sin 2θ(√(h(h cos^2 2θ+a))) such that the ball hits the plane again, x=((2h sin^2 2θ)/(tan 2θ))+2 sin 2θ(√(h(h cos^2 2θ+a)))≤b=(a/(tan θ)) ((2 sin^2 2θ)/(tan 2θ))+2 sin 2θ(√(cos^2 2θ+(a/h)))≤(a/(h tan θ)) let λ=(a/h) ⇒cos 2θ+(√(cos^2 2θ+λ))≤(λ/(2(1−cos 2θ))) example: θ=30° λ=2.2873 i.e. when (a/h)≥2.2873, the ball will hit the plane again after the impact.

$${say}\:{impact}\:{point}\:{B}\:{is}\:{a}\:{above}\:{the} \\ $$$${ground}. \\ $$$${speed}\:{at}\:{point}\:{B}: \\ $$$${v}=\sqrt{\mathrm{2}{gh}} \\ $$$${upon}\:{point}\:{B}: \\ $$$${x}={v}\:\mathrm{sin}\:\mathrm{2}\theta\:{t} \\ $$$${y}={a}+{v}\:\mathrm{cos}\:\mathrm{2}\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${y}={a}+{v}\:\mathrm{cos}\:\mathrm{2}\theta×\frac{{x}}{{v}\:\mathrm{sin}\:\mathrm{2}\theta}−\frac{{g}}{\mathrm{2}}×\left(\frac{{x}}{{v}\:\mathrm{sin}\:\mathrm{2}\theta}\right)^{\mathrm{2}} \\ $$$${y}={a}+\frac{{x}}{\mathrm{tan}\:\mathrm{2}\theta}−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta} \\ $$$$\Rightarrow{y}={a}+\frac{{x}}{\mathrm{tan}\:\mathrm{2}\theta}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}{h}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta} \\ $$$${at}\:{y}=\mathrm{0}: \\ $$$${a}+\frac{{x}}{\mathrm{tan}\:\mathrm{2}\theta}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}{h}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{4}{h}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta{x}}{\mathrm{tan}\:\mathrm{2}\theta}−\mathrm{4}{ah}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}{h}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta}{\mathrm{tan}\:\mathrm{2}\theta}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta\sqrt{{h}\left({h}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta+{a}\right)} \\ $$$${such}\:{that}\:{the}\:{ball}\:{hits}\:{the}\:{plane}\:{again}, \\ $$$${x}=\frac{\mathrm{2}{h}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta}{\mathrm{tan}\:\mathrm{2}\theta}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta\sqrt{{h}\left({h}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta+{a}\right)}\leqslant{b}=\frac{{a}}{\mathrm{tan}\:\theta} \\ $$$$\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta}{\mathrm{tan}\:\mathrm{2}\theta}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta\sqrt{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta+\frac{{a}}{{h}}}\leqslant\frac{{a}}{{h}\:\mathrm{tan}\:\theta} \\ $$$${let}\:\lambda=\frac{{a}}{{h}} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}\theta+\sqrt{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta+\lambda}\leqslant\frac{\lambda}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)} \\ $$$${example}:\:\theta=\mathrm{30}° \\ $$$$\lambda=\mathrm{2}.\mathrm{2873} \\ $$$${i}.{e}.\:{when}\:\frac{{a}}{{h}}\geqslant\mathrm{2}.\mathrm{2873},\:{the}\:{ball}\:{will} \\ $$$${hit}\:{the}\:{plane}\:{again}\:{after}\:{the}\:{impact}. \\ $$

Commented by Tawa11 last updated on 14/May/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com