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NEET Physics QUESTION #7326
Question 1

A ball with kinetic energy $E$ is projected at $45°$ to the horizontal. Its kinetic energy at the highest point of its flight is:

  • $E$
  • $\dfrac{E}{\sqrt{2}}$
  • $\dfrac{E}{2}$✔️
  • Zero
Correct Answer Explanation

At the highest point, only horizontal velocity remains: $v_x = u\cos45° = \dfrac{u}{\sqrt{2}}$

$KE_{\text{top}} = \dfrac{1}{2}mv_x^2 = \dfrac{1}{2}m\cdot\dfrac{u^2}{2} = \dfrac{1}{2}\cdot\dfrac{1}{2}mu^2 = \mathbf{\dfrac{E}{2}}$