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
The second ball velocity is larger than the first ball because second has more height than the first ball. Option 2 is correct.
From the equations of motion,
The second ball was thrown up , on getting to its highest point V=0,
[tex]\bold {H = \dfrac 12gt^2}[/tex]
Where,
H - height
g - gravitational acceleration
t - time
The second ball has two heights, one the distance from the rooftop and the another that it reached when the ball thrown up.
So, Velocity of the second ball,
[tex]\bold {v^2= u^2+2a(H1+H2)}[/tex])
Velocity of second ball,
[tex]\bold { v^2=2gH}[/tex]
Therefore, the second ball velocity is larger than the first ball because second has more height than the first ball.
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