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} $
