First slide

Projection Under uniform Acceleration

Question

A projectile is fired at an angle of $45^{0}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

Moderate

A

$60^{0}$
B

$\tan^{- 1} (\frac{1}{2})$
C

$\tan^{- 1} (\frac{\sqrt{3}}{2})$
D

$45^{0}$

Solution

Height of projectile
$H = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{u^{2} \sin^{2} 45^{0}}{2 g}$
$\Rightarrow H = \frac{u^{2}}{4 g} - - - - - - (i)$
Range of projectile R = $\frac{u^{2} \sin 2 θ}{g} = \frac{u^{2} \sin 90^{0}}{g}$
$\Rightarrow R = \frac{u^{2}}{g}$
$\tan α = \frac{H}{\frac{R}{2}}$
In $∆ OAB = \frac{\frac{u^{2}}{4 g}}{\frac{u^{2}}{2 g}}$
tan $α = \frac{1}{2} \Rightarrow α = \tan^{- 1} (\frac{1}{2})$

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