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

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Correct answer: B. tanα - 5secα

A projectile is launched at an angle α with the horizontal with a velocity 20 ms ^ - 1 . After 10s , its inclination

Component of velocity in horizontal direction,

${v}_{x} = 20cos\alpha = {v}_{2}cos\beta$

Component of velocity in vertical direction,

${v}_{y} = 20sin\alpha - g \times 10 = {v}_{2}sin\beta$

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

$⇒ tan\beta = tan\alpha - \dfrac{5}{cos\alpha}$

$⇒ tan\beta = tan\alpha - 5sec\alpha$

A projectile is launched at an angle α with the horizontal with a...

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