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Question Number 71054 by jagannath19 last updated on 11/Oct/19

Commented by jagannath19 last updated on 11/Oct/19

$$withexplanation$$

Commented by mr W last updated on 11/Oct/19

$$x=v\cos \theta t$$$$y=v\sin \theta t-\frac{1}{2}{gt}^2$$$$t=\frac{v\sin \theta }{g}$$$$y_{max}=\frac{v^2{\sin }^2\theta }{2g}=\frac{v^2}{4g}=100$$$$\Rightarrow \frac{v^2}{g}=400$$$$v\cos \theta t=300$$$$t=\frac{300}{v\cos \theta }$$$$y=t(v\sin \theta -\frac{1}{2}gt)=\frac{300}{v\cos \theta }(v\sin \theta -\frac{300g}{2v\cos \theta })$$$$=300(\tan \theta -\frac{300g}{2v^2{\cos }^2\theta })$$$$=300(1-\frac{300}{400})$$$$=75m=heightofball$$

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