Question

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-

kg ball with zero net charge was dropped from rest at a height of 1.00 m. The ball landed 0.450s later. Next, the ball was given a net charge of 7.80?C and dropped in the same way from the same height. This time the ball fell for 0.635s before landing.

What is the electric potential at a height of 1.00 m above the ground on this planet, given that the electric potential at ground level is zero? (Air resistance can be ignored.)

Answer 1

h = 0 + 0.5 gt^2 >> when no air resistance is considered
t = [1 /0.5*9.8]^1/2 = 0.452 s
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free fall time is given as 0.588 s
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2 times of fall (calculated and given by you) differ

air resistance CAN NOT BE IGNORED
otherwise you do not give t = 0.588
we will take it as = 0.452 s
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means air resistance was considered >
h = 0.5 (a) t^2
where (a) is net acceleration which discounts air resistance and gravity as counter-active
a = 1/0.5*(0.588)^2 = 5.785 m/s^2
a > net down wards acceleration
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ball is charged (Q = 7.25*10^-6 C)
dropped from same height
time take = t (field) = 0.680 s > more than 0.588 s
inference > electric field (E) opposed the NOT-SO-FREE Fall further
E > acting counter to gravity as well
modified [net] acceleration (a1) = a - (EQ/m)
a1 = a - (EQ/m) ------------ (1)
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ground potential V(g) =0
V(top) = V
E = = - dV/dx = - [v(g) - V(top)] / h = V / h
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a1 = a - (V*Q/m*h)
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h = 0 + 0.5 {a1} t(field)^2
2 h = {a - (V*Q/m*h)} 0.68*0.68
2*1/0.68*0.68 = {5.785 - (V*Q/m*1)}
4.325 = {5.785 - (V*Q/m*1)}
V*Q/m = 5.785 - 4.325 = 1.46
V (top) = 1.46*m/Q = 1.46*0.250/7.25*10^-6
V (top) = 50344.83 volts
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you must respond >> by editing
why 2 times of free fall R different