First slide

Projection Under uniform Acceleration

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

Moderate

A

$θ$
B

$\tan^{- 1} (θ / t)$
C

$\tan^{- 1} (\frac{Vcos θ}{Vsin θ - gt})$
D

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

Solution

See fig. (5). Here $v^{'} \cos α = vcos θ$
and $v^{'} \sin α = vsin θ - gt$
$\begin{array}{l} ∴ \frac{V^{'} \sin α}{V^{'} \cos α} = \frac{Vsin θ - gt}{Vcos θ} \\ or \tan α = \frac{Vsin θ - gt}{Vcos θ} \\ α = \tan^{- 1} [\frac{(Vsin θ - gt)}{Vcos θ}] \end{array}$

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