Kinematics Question 697

Question: A ball is projected with kinetic energy $ E $ at an angle of $ 45^{o} $ to the horizontal. At the highest point during it’s flight, it’s kinetic energy will be

Options:

A) Zero

B) $ \frac{E}{2} $

C) $ \frac{E}{\sqrt{2}} $

D) $ E $

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $ E’=E{{\cos }^{2}}\theta =E{{\cos }^{2}}(45{}^\circ )=\frac{E}{2} $



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