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NEET PHYSICSEasy

A ball is projected with kinetic energy EE at an angle of 45∘45^\circ to the horizontal. At the highest point during its flight, its kinetic energy will be:

A

Zero

B

E/2E/2

C

E/2E/\sqrt{2}

D

EE

Step-by-Step Solution

  1. Initial Energy: The initial kinetic energy is given by E=12mv2E = \frac{1}{2}mv^2, where vv is the initial speed of projection .
  2. Velocity at Highest Point: In projectile motion, the horizontal component of velocity (vx=vcos⁡θv_x = v \cos \theta) remains constant, while the vertical component (vyv_y) becomes zero at the highest point. Thus, the velocity at the highest point is v′=vcos⁡45∘=v2v' = v \cos 45^\circ = \frac{v}{\sqrt{2}} .
  3. Final Energy: The kinetic energy at the highest point is: E′=12m(v′)2=12m(v2)2=12m(v22)=12(12mv2)E' = \frac{1}{2}m(v')^2 = \frac{1}{2}m \left(\frac{v}{\sqrt{2}}\right)^2 = \frac{1}{2}m \left(\frac{v^2}{2}\right) = \frac{1}{2} \left( \frac{1}{2}mv^2 \right).
  4. Conclusion: Substituting EE, we get E′=E/2E' = E/2.
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