Answer:
2 the speed of the second ball is larger than that of the first one
Explanation:
Using the equations of motion
The second ball was thrown up , on getting to its highest point V=0, H=1/2gt^2
The second ball was. Now has two heights , the distance from the rooftop and the height it reached when the ball was thrown up.
v^2= u^2+2a(H1+H2)
The first ball has a velocity of v^2=2gH
Therefore the second ball velocity is larger than the first ball
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We have that The ratio U1/U2 of their potential energies due to their interactions with Q is
From the question we are told that
Question 1
Charge q1 is distance r from a positive point charge Q.
Question 2
Charge q2=q1/3 is distance 2r from Q.
Charge q1 is distance s from the negative plate of a parallel-plate capacitor.
Charge q2=q1/3 is distance 2s from the negative plate.
Generally the equation for the potential energy is mathematically given as

Therefore
The Equations of U1 and U2 is
For U1

For U2

Since
U is a function of q and q2=q1/3
Therefore

For Question 2
For U1

Therefore

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