HELP IS NEEDED ASAP<3 Particles q1, 92, and q3 are in a straight line.
Particles
91 =
= -28.1 uC, 92 = +25.5 uc, and
q3 = -47.9 uc. Particles q1 and q2 are separated
by 0.300 m. Particles q2 and q3 are separated by
0.300 m. What is the net force on q3?

Divide 108.2 x 10^6 into 1.5 x 10 and the result will be in au (astronomical units)

Ais the answer hope this

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