Using sources that are peer reviewed
Answer:
18 m
Explanation:
E = mgh
45 J = (0.25 kg) (9.8 m/s²) h
h = 18 m
In BPC
tan\theta =a/b = 3/4
\theta = tan^-1(0.75)
\theta = 36.87 deg
BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m
Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C
Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C
Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C
Net electric field along X-direction is given as
Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C
Net electric field along X-direction is given as
Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C
Net electric field is given as
E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
