Solution:
During Projectile motion horizontal component of velocity remains same so,
\[ u cos\alpha = v cos \beta \]
\[ \frac{u cos\alpha}{cos \beta} = v \]
\[ {u .cos\alpha}.{sec \beta} = v \]
A particle is thrown with the speed u at an angle α with the horizontal. When the particle makes an angle β with the horizontal, its speed will be :
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