First slide

Projectile motion

Question

A particle is projected with a certain velocity at an angle $α$ above the horizontal from the foot of an inclined plane of inclination 30^o. If the particle strikes the plane normally, then $α$ is equal to

Difficult

A

$30^{\circ} + \tan^{- 1} (\frac{\sqrt{3}}{2})$
B

$45^{\circ}$
C

$60^{\circ}$
D

$30^{\circ} + \tan^{- 1} (2 \sqrt{3})$

Solution

$t_{A B} = time of flight of projectile$
$= \frac{2 u \sin (α - 30^{\circ})}{g \cos 30^{\circ}}$
Now the component of velocity along the plane becomes zero at point B.
$\begin{array}{l} θ = u \cos (α - 30^{\circ}) - g \sin 30^{\circ} \times t_{A B} \\ o r u \cos (α - 30^{\circ}) = g \sin 30^{\circ} \times \frac{2 u \sin (α - 30^{\circ})}{g \cos 30^{\circ}} \\ o r \tan (α - 30^{\circ}) = \frac{\cot 30^{\circ}}{2} = \frac{\sqrt{3}}{2} \end{array}$
$α = 30^{\circ} + \tan^{- 1} (\frac{\sqrt{3}}{2})$

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