The electric field midway between two equal but opposite point charges is 386 n/c, and the distance between the charges is 16.0 cm. what is the magnitude of the charge on each?

We'll use the following formular E = Kq / r^2 Since the electric field is 386 N/C midway between the charges, r = 1/2 * 16 = 8 cm = 0.08m. So we have 386 = (8.998 * 10^(9) * q) / (0.08)^2 386 * 0.0064 = 8.998 * 10^(9) * q q = (386 * 0.0064)/ 8.998 * 10^(9) q = 2.4704/ 8.998 * 10^(9) q = 2.754 * 10^(-10)

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