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Which one of the following gives the angle made by the resultant velocity after t seconds for projectile of initial velocity u, projected at an angle $θ$ with the horizontal ?

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$\tan^{- 1} (\frac{gt}{ucos θ})$

$\tan^{- 1} (\frac{t}{ucos θ})$

$\tan^{- 1} (\tan θ - \frac{gt}{ucos θ})$

$\tan^{- 1} (\tan θ + \frac{gt}{ucos θ})$

answer is C.

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Detailed Solution

Here $v_{x} = ucos θ and v_{y} = usin θ - gt$

$\begin{array}{l} ∴ \tan α = \frac{V_{y}}{V_{x}} = \frac{usin θ - gt}{ucos θ} \\ = (\tan θ - \frac{gt}{ucos θ}) \\ ∴ α = \tan^{- 1} (\tan θ - \frac{gt}{ucos θ}) \end{array}$

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