Let's start by finding the final vertical velocity of Ball A that was launched vertically. We know that the ball was launched from an initial height of 3m with initial vertical velocity V.

Given that, we know that its initial mechanical energy is

`E_i = mgH + 1/2 mV_y_B^2 `

`E_i = (2 kg)(10 m/(s^2))(3 m) + 1/2 (2 kg)(4 m/s)² = (60 J) + (16 J) = 76 J `

And we know that the mechanical energy when it hits the ground (assuming no air resistance) will be

`E_f = mg0 + 1/2mVf_y_A^2`

`E_f = 0 J + 1/2 (2 kg) Vf_y_A^2 = Vf_y_A^2`

Since in our case the conservation of mechanical energy is valid, we have Ei = Ef, so we have

`Vf_y_A^2 = 76 J`

`Vf_y_A = 8.7 m/s`

This is the final velocity of Ball A, because there is no horizontal component to it!

Now, let's solve for the ball that was launched horizontally. Again, we will use the conservation of mechanical energy to find the final vertical velocity.

The initial energy will be

`E_i = mgH + 1/2mV_y_B^2`

`E_i = (2 kg)(10 m/(s^2))(3 m) + 1/2 (2 kg)(0 m/s)² = 60 J`

Note that in this case, the vertical velocity is 0. Now, solving for the final vertical velocity, using the final mechanical energy

`E_f = mg0 + 1/2(2 kg) Vf_y_B^2`

`E_f = Vf_y_B²`

Setting Ei = Ef give us

`Vf_y_B^2 = 60 J`

`Vf_y_B² = 7.75 m/s`

But in this case, we also have a horizontal component to the velocity. Since we assumed no air resistance, the final horizontal velocity will be equal to the initial horizontal velocity; that is,

`Vf_x_B = 3m/s`

Using the Pythagorean Theorem, we have

`Vf_B^2 = (7.75 m/s)^2 + (3 m/s) ^2 = 60 + 9 = 69`

`Vf_B = 8 m/s`

Now, we simply take the ratio of `Vf_A` and `Vf_B`

`(Vf_A) / (Vf_B) = 8.72/8 = 1.09`

Thus, Ball A's final velocity is 1.09 times Ball B's final velocity.

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