Q64 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
An object is projected with kinetic energy $K$ from a point $A$ at an angle $60^{\circ}$ with the horizontal. The ratio of the difference in kinetic energies at points $B$ and $C$ to that at point $A$ (see figure), in the absence of air friction is :
- A$1: 4$
- B$2: 3$
- Correct answerC$3: 4$
- D$1: 2$
Solution
At A: $KE_A = K$, $v_A = v$, angle $= 60°$.
At B (highest point): $v_B = v\cos 60° = v/2$, so $KE_B = K/4$.
At C (same level as A): by energy conservation, $KE_C = K$.
$|KE_B - KE_C| = |K/4 - K| = 3K/4$.
Ratio $= \frac{3K/4}{K} = \frac{3}{4}$, i.e., $3:4$.
At B (highest point): $v_B = v\cos 60° = v/2$, so $KE_B = K/4$.
At C (same level as A): by energy conservation, $KE_C = K$.
$|KE_B - KE_C| = |K/4 - K| = 3K/4$.
Ratio $= \frac{3K/4}{K} = \frac{3}{4}$, i.e., $3:4$.
Hence, the correct answer is option: 'C'.