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Show that projection angle theta(0) for ...

Show that projection angle `theta_(0)` for a projectile launched from the origin is given by:
`theta_(0)=tan^(-1)[(4H_(m))/(R)]` Where `H_(m)=` Maximum height, R=Range

Text Solution

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`H_(m)=(u^(2)sin^(2)theta_(0))/(2g)` and `R=(2u^(2))/(g)xxsintheta_(0)xxcostheta_(0)`
So, `(H_(m))/(R)=(tantheta_(0))/(4)rArrtantheta_(0)=(4H_(m))/(R)rArr=tan^(-1)((4H_(m))/(R))`
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