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Question Number 177055 by ajfour last updated on 30/Sep/22

Commented by ajfour last updated on 30/Sep/22

Our boy throws a ball with speed u and starts climbing down at same speed u and runs forward to catch the ball, and does so success -fully. Find θ in terms of u,g,h.

$${Our}\:{boy}\:{throws}\:{a}\:{ball}\:{with}\:{speed}\:{u} \\ $$$${and}\:{starts}\:{climbing}\:{down}\:{at}\:{same} \\ $$$${speed}\:{u}\:{and}\:{runs}\:{forward}\:{to}\: \\ $$$${catch}\:{the}\:{ball},\:{and}\:{does}\:{so}\:{success} \\ $$$$-{fully}.\:{Find}\:\theta\:{in}\:{terms}\:{of}\:{u},{g},{h}. \\ $$

Commented by peter frank last updated on 30/Sep/22

how do you draw this?

$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{draw}\:\mathrm{this}? \\ $$

Commented by mr W last updated on 01/Oct/22

every body can draw with his smart phone. also i can do :). but whether you can draw as nicely as ajfour sir, is an other question.

$${every}\:{body}\:{can}\:{draw}\:{with}\:{his}\:{smart} \\ $$$$\left.{phone}.\:{also}\:{i}\:{can}\:{do}\::\right).\:\:{but}\:{whether} \\ $$$${you}\:{can}\:{draw}\:{as}\:{nicely}\:{as}\:{ajfour}\:{sir},\: \\ $$$${is}\:{an}\:{other}\:{question}. \\ $$

Commented by mr W last updated on 01/Oct/22

Commented by peter frank last updated on 01/Oct/22

what the name of the app

$$\mathrm{what}\:\mathrm{the}\:\mathrm{name}\:\mathrm{of}\:\mathrm{the}\:\mathrm{app} \\ $$

Commented by mr W last updated on 02/Oct/22

every smart phone has an app for drawing/editing pictures! the name of the app doesn′t matter.

$${every}\:{smart}\:{phone}\:{has}\:{an}\:{app}\:{for} \\ $$$${drawing}/{editing}\:{pictures}!\:{the}\:{name} \\ $$$${of}\:{the}\:{app}\:{doesn}'{t}\:{matter}. \\ $$

Answered by mr W last updated on 30/Sep/22

ut=h+u cos θ t ⇒t=(h/(u(1−cos θ))) u sin θ t−((gt^2 )/2)=−h u sin θ×(h/(u(1−cos θ)))−(g/2)×((h/(u(1−cos θ))))^2 =−h sin θ+1−cos θ−((gh)/(2u^2 (1−cos θ)))=0 sin (θ/2) cos (θ/2)+sin^2 (θ/2)−((gh)/(8u^2 sin^2 (θ/2)))=0 with λ=((gh)/(8u^2 )) sin^3 (θ/2) cos (θ/2)+sin^4 (θ/2)−λ=0 sin^3 (θ/2) cos (θ/2)=λ−sin^4 (θ/2) sin^6 (θ/2)(1−sin^2 (θ/2))=λ^2 −2λ sin^4 (θ/2)+sin^8 (θ/2) ⇒2 sin^8 (θ/2)−sin^6 (θ/2)−2λ sin^4 (θ/2)−λ^2 =0 example: u=5 m/s, h=5m ⇒λ=((10×5)/(8×5^2 ))=(1/4) ⇒θ≈68.7325°

$${ut}={h}+{u}\:\mathrm{cos}\:\theta\:{t} \\ $$$$\Rightarrow{t}=\frac{{h}}{{u}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)} \\ $$$${u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=−{h} \\ $$$${u}\:\mathrm{sin}\:\theta×\frac{{h}}{{u}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}−\frac{{g}}{\mathrm{2}}×\left(\frac{{h}}{{u}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}\right)^{\mathrm{2}} =−{h} \\ $$$$\mathrm{sin}\:\theta+\mathrm{1}−\mathrm{cos}\:\theta−\frac{{gh}}{\mathrm{2}{u}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\theta\right)}=\mathrm{0} \\ $$$$\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}+\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}−\frac{{gh}}{\mathrm{8}{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}=\mathrm{0} \\ $$$${with}\:\lambda=\frac{{gh}}{\mathrm{8}{u}^{\mathrm{2}} } \\ $$$$\mathrm{sin}^{\mathrm{3}} \:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}+\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}−\lambda=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{3}} \:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\lambda−\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{6}} \:\frac{\theta}{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)=\lambda^{\mathrm{2}} −\mathrm{2}\lambda\:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}+\mathrm{sin}^{\mathrm{8}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\:\mathrm{sin}^{\mathrm{8}} \:\frac{\theta}{\mathrm{2}}−\mathrm{sin}^{\mathrm{6}} \:\frac{\theta}{\mathrm{2}}−\mathrm{2}\lambda\:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}−\lambda^{\mathrm{2}} =\mathrm{0} \\ $$$${example}: \\ $$$${u}=\mathrm{5}\:{m}/{s},\:{h}=\mathrm{5}{m}\:\Rightarrow\lambda=\frac{\mathrm{10}×\mathrm{5}}{\mathrm{8}×\mathrm{5}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\theta\approx\mathrm{68}.\mathrm{7325}° \\ $$

Commented by ajfour last updated on 30/Sep/22

Thanks sir, dont seem a simpler way out..

$${Thanks}\:{sir},\:{dont}\:{seem}\:{a}\:{simpler} \\ $$$${way}\:{out}.. \\ $$

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