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Question

$	heta , 90^o -	heta $

${y_1} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}$ and ${y_2} = \frac{{{u^2}{{\cos }^2}	heta }}{{2g}}$

$	herefore {y_1}

Vedclass · Accepted Answer

$\frac{{{u^2}}}{{2g}}$

Vedclass · Answer

$	heta , 90^o -	heta $

${y_1} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}$ and ${y_2} = \frac{{{u^2}{{\cos }^2}	heta }}{{2g}}$

$	herefore {y_1} + {y_2} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}} + \frac{{{u^2}{{\cos }^2}	heta }}{{2g}}$

$	herefore {y_1} + {y_2} = \frac{{{u^2}}}{{2g}}$

A ball is thrown at different angles with the same speed $u$ and from the same point. It has the same range in both cases. If $y_1$ and $y_2$ be the heights attained in the two cases, then $y_1+y_2$ equals to

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