Kinematics Question 530
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 its point of projection and highest point of trajectory is
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 } $
BETA
