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Question

The angle which the velocity vector of a projectile, thrown with a velocity $v$ at an angle $θ$ to the horizontal, will make with the horizontal after time $t$ of its being thrown up is
$θ$
$t a n^{- 1} (\frac{θ}{t})$
$t a n^{- 1} (\frac{v c o s θ}{v s i n θ - g t})$
$t a n^{- 1} (\frac{v s i n θ - g t}{v c o s θ})$

A

t a n^{- 1} (\frac{v c o s θ}{v s i n θ - g t})

B

θ

C

t a n^{- 1} (\frac{v s i n θ - g t}{v c o s θ})

D

t a n^{- 1} (\frac{θ}{t})

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Solution

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Horizontal velocity of projectile after time t $= v c o s θ$
Vertical velocity of projectile after time t $= v s i n θ - g t$
Angle at any time t:
$tan ϕ = \frac{v_{v}}{v_{h}} = \frac{v s i n θ - g t}{v c o s θ} ϕ = {t a n}^{- 1} \frac{v s i n θ - g t}{v c o s θ}$

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