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}} $
Show Answer
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} $
