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Work Power and Energy question

2002 · Shift 0 · Q168

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Work Power and Energy
2002 · Shift 0 · Q168

Work Power and Energy question

2002 · Shift 0 · Q168

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1

A ball whose kinetic energy E, is projected at an angle of

45^\circ

to the horizontal. The kinetic energy of the ball at the highest point of its height will be

A
E
B
${E \over {\sqrt 2 }}$
C
${E \over 2}$
D
zero

View written solutionFree

Correct answer: C

Initial kinetic energy

Let the ball be projected with speed $u$ .

Then its initial kinetic energy is

E = \frac{1}{2}mu^2

Resolve velocity into components

Since the angle of projection is $45^\circ$ ,

u_x = u\cos45^\circ = \frac{u}{\sqrt{2}}, \qquad u_y = u\sin45^\circ = \frac{u}{\sqrt{2}}

Velocity at the highest point

At the highest point of projectile motion, the vertical component of velocity becomes zero, while the horizontal component remains unchanged.

So speed at the highest point is

v = u_x = \frac{u}{\sqrt{2}}

Kinetic energy at the highest point

Therefore,

K_{\text{top}} = \frac{1}{2}m\left(\frac{u}{\sqrt{2}}\right)^2 = \frac{1}{2}m\cdot \frac{u^2}{2} = \frac{1}{4}mu^2

But

E = \frac{1}{2}mu^2

So,

K_{\text{top}} = \frac{E}{2}

Option check

A: $E$ ❌
B: $\dfrac{E}{\sqrt{2}}$ ❌
C: $\dfrac{E}{2}$ ✅
D: zero ❌

Hence, the correct answer is C.

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