Kinematics Question 606

Question: A particle of mass M is projected with a velocity u making an angle of $ 30{}^\circ $ with the horizontal. The magnitude of$ (V _{h}\times h) $ of the projectile when the particle is at it’s maximum height h

Options:

A) $ \frac{\sqrt{3}}{2}\frac{{{\text{v}}^{\text{2}}}}{\text{g}} $

B)zero

C) $ \frac{{{\text{v}}^{\text{2}}}}{\sqrt{2}\text{g}} $

D) $ \frac{\sqrt{3}}{16}\frac{{{\text{v}}^{\text{2}}}}{\text{g}} $

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Answer:

Correct Answer: D

Solution:

[d] $ {{\text{V}} _{\text{h}}}\text{= V cos}\theta $

Where h it’s the maximum height $ V _{h}\times h=( v\cos \theta)( \frac{v^{2}{{\sin }^{2}}\theta }{2g} ) $

$ =\frac{v^{3}{{\sin }^{2}}\theta \cos \theta }{2g}=\frac{\sqrt{3}v^{3}}{16g} $



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