Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 174823 by ajfour last updated on 11/Aug/22

Commented by ajfour last updated on 11/Aug/22

Find maximum height the ball  can climb up the wall.

$${Find}\:{maximum}\:{height}\:{the}\:{ball} \\ $$$${can}\:{climb}\:{up}\:{the}\:{wall}. \\ $$

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

t_1 =(b/(u cos θ))  v_1 =u sin θ−((gb)/(u cos θ)) ≥0  sin 2θ≥((2gb)/u^2 )  ⇒(1/2)sin^(−1) ((2gb)/u^2 )≤θ≤(π/2)−(1/2)sin^(−1) ((2gb)/u^2 )  ⇒u≥(√(2gb))  max. height reached:  h=(((u sin θ)^2 )/(2g))=((u^2 (1−cos 2θ))/(4g))     ≤ (u^2 /(4g))[1+(√(1−(((2gb)/u^2 ))^2 ))]=h_(max)

$${t}_{\mathrm{1}} =\frac{{b}}{{u}\:\mathrm{cos}\:\theta} \\ $$$${v}_{\mathrm{1}} ={u}\:\mathrm{sin}\:\theta−\frac{{gb}}{{u}\:\mathrm{cos}\:\theta}\:\geqslant\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{2}\theta\geqslant\frac{\mathrm{2}{gb}}{{u}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}{gb}}{{u}^{\mathrm{2}} }\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}{gb}}{{u}^{\mathrm{2}} } \\ $$$$\Rightarrow{u}\geqslant\sqrt{\mathrm{2}{gb}} \\ $$$${max}.\:{height}\:{reached}: \\ $$$${h}=\frac{\left({u}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{{u}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)}{\mathrm{4}{g}} \\ $$$$\:\:\:\leqslant\:\frac{{u}^{\mathrm{2}} }{\mathrm{4}{g}}\left[\mathrm{1}+\sqrt{\mathrm{1}−\left(\frac{\mathrm{2}{gb}}{{u}^{\mathrm{2}} }\right)^{\mathrm{2}} }\right]={h}_{{max}} \\ $$

Commented by ajfour last updated on 13/Aug/22

thanks sir, perfect!

$${thanks}\:{sir},\:{perfect}! \\ $$

Commented by Tawa11 last updated on 14/Aug/22

Great sir

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com