Work Energy And Power Question 205
Question: A ball whose kinetic energy is E, is projected at an angle of $ 45{}^\circ $ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be
Options:
A) E
B) $ E/\sqrt{2} $
C) $ E/2 $
D) zero
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let u be the speed with which the ball of mass m is projected.
Then the kinetic energy (E) at the point of projection is $ E=\frac{1}{2}mu^{2} $ -(i)
When the ball is at the highest point of its flight, the speed of the ball is $ \frac{u}{\sqrt{2}} $ (Remember that the horizontal component of velocity does not change during a projectile motion).
$ \therefore $ The kinetic energy at the highest point $ =\frac{1}{2}m{{( \frac{u}{\sqrt{2}} )}^{2}}=\frac{1}{2}\frac{mu^{2}}{2}=\frac{E}{2} $ [From (i)]
