SATHEE: Work Energy And Power Question 205

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 $E = \frac{1}{2} m u^{2}$ -(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).
$∴$ The kinetic energy at the highest point $= \frac{1}{2} m {(\frac{u}{\sqrt{2}})}^{2} = \frac{1}{2} \frac{m u^{2}}{2} = \frac{E}{2}$ [From (i)]

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Work Energy And Power Question 205

Question: A ball whose kinetic energy is E, is projected at an angle of 45∘ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

Options:

Answer:

Solution:

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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