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

Commented by ajfour last updated on 30/Sep/22

$$Ourboythrowsaballwithspeedu$$$$andstartsclimbingdownatsame$$$$speeduandrunsforwardto$$$$catchtheball,anddoessosuccess$$$$-fully.Find\theta intermsofu,g,h.$$

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

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

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

$$everybodycandrawwithhissmart$$$$phone.alsoicando:).butwhether$$$$youcandrawasnicelyasajfoursir,$$$$isanotherquestion.$$

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

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

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

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

$$everysmartphonehasanappfor$$$$drawing/editingpictures!thename$$$$oftheapp{doesn}^{\prime }tmatter.$$

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

$$ut=h+u\cos \theta t$$$$\Rightarrow t=\frac{h}{u(1-\cos \theta )}$$$$u\sin \theta t-\frac{{gt}^2}{2}=-h$$$$u\sin \theta \times \frac{h}{u(1-\cos \theta )}-\frac{g}{2}\times {\left(\frac{h}{u(1-\cos \theta )}\right)}^2=-h$$$$\sin \theta +1-\cos \theta -\frac{gh}{2u^2(1-\cos \theta )}=0$$$$\sin \frac{\theta }{2}\cos \frac{\theta }{2}+{\sin }^2\frac{\theta }{2}-\frac{gh}{8u^2{\sin }^2\frac{\theta }{2}}=0$$$$with\lambda =\frac{gh}{8u^2}$$$${\sin }^3\frac{\theta }{2}\cos \frac{\theta }{2}+{\sin }^4\frac{\theta }{2}-\lambda =0$$$${\sin }^3\frac{\theta }{2}\cos \frac{\theta }{2}=\lambda -{\sin }^4\frac{\theta }{2}$$$${\sin }^6\frac{\theta }{2}(1-{\sin }^2\frac{\theta }{2})={\lambda }^2-2\lambda {\sin }^4\frac{\theta }{2}+{\sin }^8\frac{\theta }{2}$$$$\Rightarrow 2{\sin }^8\frac{\theta }{2}-{\sin }^6\frac{\theta }{2}-2\lambda {\sin }^4\frac{\theta }{2}-{\lambda }^2=0$$$$example:$$$$u=5m/s,h=5m\Rightarrow \lambda =\frac{10\times 5}{8\times 5^2}=\frac{1}{4}$$$$\Rightarrow \theta \approx 68.7325°$$

Commented by ajfour last updated on 30/Sep/22

$$Thankssir,dontseemasimpler$$$$wayout..$$

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