Question

A projectile is launched at an angle ' $\alpha$ ' with the horizontal with a velocity $20\, ms ^{-1}$. After $10 \,s$, 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. $2 \tan \alpha+5 \sec \alpha$

Answer 1

Answer: (B)

$v_{x}=u_{x}=20 \cos \alpha$
$v_{y}=20 \sin \alpha-10 \times 10$
$\tan \beta=\frac{v_{y}}{v_{x}}=\frac{20 \sin \alpha-100}{20 \cos \alpha}$
$=\tan \alpha-5 \sec \alpha$