Kinematics Question 530

Question: A particle is projected from the ground with an initial speed of $ v $ at an angle $ \theta $ with horizontal. The average velocity of the particle between it’s point of projection and highest point of trajectory it’s

Options:

A) $ \frac{v}{2}\sqrt{1+2{{\cos }^{2}}\theta } $

B) $ \frac{v}{2}\sqrt{1+2{{\cos }^{2}}\theta } $

C) $ \frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta } $

D) $ v\cos \theta $

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Answer:

Correct Answer: C

Solution:

[c] Average velocity $ =\frac{Displacement}{Time} $

$ =\frac{\sqrt{H^{2}+R^{2}/4}}{T/2} $

Putting the required values, we get $ v _{av}=\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta } $



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