# physics

On planet Seeturn, the free fall acceleration is 9.800 N/kg but there is also a strong downward electric field that is uniform close to the planet's surface. A 2.00 kg ball having a charge of 5.00 micro C (5.00*10-6) is thrown upward at a speed of 20.1 m/s, and it hits the ground after an interval of 4.10s. What is the potential difference between the starting point and the top point of the trajectory?

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1. What is the kinetic energy at the bottom?
Ke = (1/2) m v^2 = .5 * 2 * 20.1^2

What is the average speed going up?
Vav = 20.1 / 2 = 10.05 m/s

So how long did it take to go up?
t = 4.1/2 = 2.05 seconds upward

so how high did it go?
h = 10.05 * 2.05

so how much work did gravity do on the way up?
Wg = m g h = 2*9.8* h

so how much was used to climb the voltage potential
Ke - Wg

but we all know that is the charge times the voltage (the potential difference)
Q V = Ke -Wg

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posted by Damon
2. Bob Pursley already did this for you.

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posted by Damon
3. What would be the final answer and what would be the value of Q

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posted by john
4. 5.00 micro C (5.00*10-6)
is Q, the charge. It is GIVEN.
No, you do it.

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posted by Damon

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posted by paul

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