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(B) The initial velocity components are $u_x = 20 \cos \alpha$ and $u_y = 20 \sin \alpha$.Since the horizontal acceleration is zero,the horizontal vel

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$	an \alpha - 5 \sec \alpha$

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(B) The initial velocity components are $u_x = 20 \cos \alpha$ and $u_y = 20 \sin \alpha$.Since the horizontal acceleration is zero,the horizontal velocity remains constant: $v_x = u_x = 20 \cos \alpha$.The vertical velocity after time $t = 10 \; s$ is given by $v_y = u_y - gt = 20 \sin \alpha - 10 	imes 10 = 20 \sin \alpha - 100$.The inclination $\beta$ with the horizontal is given by $	an \beta = \frac{v_y}{v_x}$.Substituting the values: $	an \beta = \frac{20 \sin \alpha - 100}{20 \cos \alpha} = \frac{20 \sin \alpha}{20 \cos \alpha} - \frac{100}{20 \cos \alpha} = 	an \alpha - 5 \sec \alpha$.

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

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