one person is moving towards east and another is moving along the path bisecting north-east with same velocity. Then how can you write the velocity of second person with repsect to first one

The answer is a i think

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

Answerthe force between the charges will be  f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes  two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges:     k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then