Question 244 of 308

A projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20{ms}^{-1}$. After $10s$, its inclination with horizontal is $\beta$. The value of $\tan \beta$ will be : $\left(g=10{ms}^{-2}\right)$.

A$\tan \alpha +5\sec \alpha$B$\tan \alpha -5\sec \alpha$C$2\tan \alpha -5\sec \alpha$D$2\tan \alpha +5\sec \alpha$

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Answer: (B) $\tan \alpha -5\sec \alpha$

Component of velocity in horizontal direction,
${v}_{x}=20\cos \alpha ={v}_{2}\cos \beta$
Component of velocity in vertical direction,
${v}_{y}=20\sin \alpha -g\times 10={v}_{2}\sin \beta$
$\Rightarrow \tan \beta =\dfrac{20\sin \alpha -100}{20\cos \alpha }$
$\Rightarrow \tan \beta =\tan \alpha -\dfrac{5}{\cos \alpha }$
$\Rightarrow \tan \beta =\tan \alpha -5\sec \alpha$