JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

An object is projected with kinetic energy K from a point A at an angle $6 0^{\circ}$ with the horizontal. The ratio of the difference in kinetic energies points B and C to that at point A (see figure), in the absence of air friction is :

A
$1 : 2$
B
$2 : 3$
C
$1 : 4$
D
$3 : 4$

Solution

Let launch speed be $u$ . Then $(K E)_{A} = \frac{1}{2} m u^{2} = K$ .

At the apex B, only the horizontal component $u cos 6 0^{\circ} = u /2$ survives:

$(K E)_{B} = \frac{1}{2} m (u /2)^{2} = \frac{K}{4}$

At C the projectile lands at the same height as A, so by energy conservation $(K E)_{C} = K$ .

$\frac{( K E ) _{C} - ( K E ) _{B}}{( K E ) _{A}} = \frac{K - K /4}{K} = \frac{3}{4}$

$∴$ ratio is $3 : 4$ .

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