SATHEE: Kinematics Question 516

Kinematics Question 516

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 $30^{\circ}$ . If the particle strikes the plane normally, then $α$ is equal to

Options:

A) $30^{\circ} + t e n^{- 1} (\frac{\sqrt{3}}{2})$

B) $45^{\circ}$

C) $60^{\circ}$

D) $30^{\circ} + t e n^{- 1} (2 \sqrt{3})$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $t_{A B}$ =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.

$θ = u \cos (α - 30^{\circ}) - g s i n 30^{\circ} \times t_{A B}$

Or $u \cos (α - 30^{\circ}) = g s i n 30^{\circ} \times \frac{2 u \sin (α - 30^{\circ})}{g \cos 30^{\circ}}$ Or $\tan (α - 30^{\circ}) = \frac{\cot 30^{\circ}}{2} = \frac{\sqrt{3}}{2}$

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

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Kinematics Question 516

Question: A particle is projected with a certain velocity at an angle α above the horizontal from the foot of an inclined plane of inclination30∘ . If the particle strikes the plane normally, then α is equal to

Options:

Answer:

Solution:

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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 $30^{\circ}$ . If the particle strikes the plane normally, then $α$ is equal to