## To solve this problem, we need to find the initial velocity (Vo) of the second ball.

The first ball reaches the same height as the initial position after 3 seconds. We can find the vertical speed of the first ball using the equation:

hf = hi + vi*t + (1/2)g*t^2

where hf is the final height (which is equal to the initial height, hi), vi is the initial velocity (which we want to find), t is the time (which is 3 seconds), and g is the acceleration due to gravity (-9.8 m/s^2).

Substituting the known values into the equation:

0 = 0 + vi*3 + (1/2)(-9.8)(3^2)

Simplifying the equation:

0 = 3vi - 44.1

Rearranging the equation to solve for vi:

vi = 44.1/3

Therefore, the vertical speed of the first ball is approximately 14.7 m/s.

Now, let's consider the second ball. The vertical component of its initial velocity (Vo) can be found using the equation for the vertical component of velocity:

Vi = Vo * sin(Î¸)

where Vi is the vertical component of velocity (which we found to be 14.7 m/s from the first ball), and Î¸ is the angle at which the second ball is thrown (which is 30 degrees).

Substituting the known values into the equation:

14.7 = Vo * sin(30)

Simplifying the equation:

14.7 = (Vo * âˆš3) / 2

Multiplying both sides by 2/âˆš3:

Vo = 2 * 14.7 / âˆš3

Therefore, the second ball must be thrown at a speed of approximately 17.0 m/s.