(c) $v = \sqrt{(8)^{2} + (6)^{2}} = 10ms^{- 1}$ $KE\ = \frac{1}{2}mv^{2}$ $= \frac{1}{2} \times 0.4 \times 10 \times 10 = 20$ J
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.