Verified Solution • 2025–26 Session moderate

QJEEM_2025_2: JEE Main 2025 JEEM_2025_2 Solution: A ball of mass 100 g is projected with velocity 20 m/s at 60 ∘…

A ball of mass 100 g is projected with velocity 20 m/s at 60 ∘ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is

Multiple Choice Options

  • A 20J
  • B 5J
  • C 15J
  • D Zero

Step-by-Step Verified Solution

Ki=12mv2K_i = \frac{1}{2} m v^2
Kf=12m(vcos⁡60∘)2=12m(v2cos⁡260∘)=12m(v2⋅14)=18mv2K_f = \frac{1}{2} m (v \cos 60^\circ)^2 = \frac{1}{2} m \left( v^2 \cos^2 60^\circ \right) = \frac{1}{2} m \left( v^2 \cdot \frac{1}{4} \right) = \frac{1}{8} m v^2
ΔK=Ki−Kf=12mv2−18mv2=38mv2\Delta K = K_i - K_f = \frac{1}{2} m v^2 - \frac{1}{8} m v^2 = \frac{3}{8} m v^2
ΔK=38×0.1×400=15 J\Delta K = \frac{3}{8} \times 0.1 \times 400 = 15\ \text{J}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring