A particle is projected from an inclined plane at an angle 37- with the inclined plane in upward direction with speed 10 m - s- The angle of inclined plane with horizontal is 53- Then the maximum height attained by the particle in the direction perpendicular to the inclined plane is- -Consider sin 37-3-5 - A- 4 mB- 5 mC- 3 mD- Zero

The correct option is C 3 mH = u^2 ⊥/2a ⊥ = (10 sin 37^∘)/2g cos 53^∘ = 3m ...

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Standard XII

Physics

Projectile Time, Height and Range

A particle is...

Question

A particle is projected from an inclined plane at an angle $37^{\circ}$ with the inclined plane in upward direction with speed 10 m/s. The angle of inclined plane with horizontal is $53^{\circ}$ . Then the maximum height attained by the particle in the direction perpendicular to the inclined plane is: [Consider $s i n 37^{\circ} = \frac{3}{5}$ ]

3 m

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5 m

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Zero

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Solution

The correct option is A 3 m
$H = \frac{u^{2} ⊥}{2 a ⊥} = \frac{(10 s i n 37^{\circ})}{2 g c o s 53^{\circ}} = 3 m$

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The correct option is A 3 mH=u2⊥2a⊥=(10 sin 37∘)2g cos 53∘=3m

The correct option is A 3 m
$H = \frac{u^{2} ⊥}{2 a ⊥} = \frac{(10 s i n 37^{\circ})}{2 g c o s 53^{\circ}} = 3 m$