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Work Energy and Power question

2001 · Q157
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Work Energy and Power question

2001 · Q157

NEETPhysicsWork Energy and PowerMCQ+4 / −1
A particle is projected making an angle of 45o with horizontal having kinetic energy K. The kinetic energy at highest point will be
  1. A
    K2{K \over {\sqrt 2 }}2​K​
  2. B
    K2{K \over 2}2K​
  3. C
    2K
  4. D
    K
View written solutionFree

Correct answer: B

Kinetic energy of the ball = K and angle of projection (θ\theta θ) = 45o

Velocity of the ball at the highest point = v cos θ\theta θ

=v  cos⁡  45o=v2 = v\,\,\cos \,\,{45^o} = {v \over {\sqrt 2 }}=vcos45o=2​v​

Therefore kinetic energy of the ball

⇒\Rightarrow⇒ 12m×(v2)=14mv2=K2{1 \over 2}m \times \left( {{v \over {\sqrt 2 }}} \right) = {1 \over 4}m{v^2} = {K \over 2}21​m×(2​v​)=41​mv2=2K​

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