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

Given: Initial velocity $$u = 20$$ m/s at angle $$\alpha$$ with horizontal, $$t = 10$$ s, $$g = 10$$ m/s$$^2$$.

The horizontal and vertical components of velocity at time $$t$$ are:

$$v_x = u\cos\alpha$$

$$v_y = u\sin\alpha - gt = 20\sin\alpha - 10 \times 10 = 20\sin\alpha - 100$$

The inclination $$\beta$$ with horizontal at time $$t = 10$$ s is:

$$\tan\beta = \frac{v_y}{v_x} = \frac{20\sin\alpha - 100}{20\cos\alpha}$$

$$\tan\beta = \frac{20\sin\alpha}{20\cos\alpha} - \frac{100}{20\cos\alpha}$$

$$\tan\beta = \tan\alpha - \frac{5}{\cos\alpha}$$

$$\tan\beta = \tan\alpha - 5\sec\alpha$$

The correct answer is Option B.