A particle is projected at 45° to the horizontal with initial kinetic energy K. What is its kinetic energy at the highest point?
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A particle is projected at 45° to the horizontal with initial kinetic energy K. What is its kinetic energy at the highest point?

G1 Physics · Accepted Answer

Step 1 — Principle: At the highest point of an ordinary projectile, the vertical velocity component is zero. Step 2 — Working: At the top vy=0 but vx=u cos45°=u/√2. Step 3 — Working: Ktop=½m(u²/2)=½(½mu²)=K/2. Correct answer: B — K/2.