Question

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

Answer 1

After $10s$,
Velocity in $y$ direction:
$V_y=20\times \sin \alpha -10\times 10$

Velocity in $x$ direction:
$V_x=20\cos \alpha$
$\therefore \tan \beta =\frac{V_y}{V_x}=\frac{20\sin \alpha -100}{20\cos \alpha }$
$=\tan \alpha -5\sec \alpha$