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Q.

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

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a

$\tan α + 5 \sec α$

b

$2 \tan α - 5 \sec α$

c

$\tan α - 5 \sec α$

d

$2 \tan α + 5 \sec α$

answer is B.

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Detailed Solution

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

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A projectile is launched at an angle ' α ' with the horizontal with a velocity 20 ms-1. After 10 s, its inclination with horizontal is ' β '. The value of tanβ will be : g=10 ms-2