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

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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)]



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