Answer:
Approximately [tex]39 \; \rm m[/tex], assuming that the gravitational field strength is [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex] and that the air resistance on this baseball is negligible.
Explanation:
Start by finding the duration [tex]t[/tex] of the flight of this baseball.
The SUVAT equation [tex]h = (1/2)\, g\, t^{2} + v_{0} \, t[/tex] relates [tex]t[/tex] to initial height [tex]h[/tex], initial vertical velocity [tex]v_{0}[/tex], and gravitational acceleration [tex]g[/tex]. (This equation applies only if the air resistance on the baseball is negligible.)
The initial vertical velocity of this baseball would be [tex]v_{0} = 0\; \rm m\cdot s^{-1}[/tex] since this ball was thrown horizontally. The equation becomes:
[tex]\begin{aligned}h &= \frac{1}{2}\, g\, t^{2} + v_{0} \, t \\ &= \frac{1}{2}\, g\, t^{2}\end{aligned}[/tex].
Rearrange and solve for [tex]t[/tex]:
[tex]\begin{aligned}t^{2} &= \frac{2\, h}{g} \end{aligned}[/tex].
[tex]\begin{aligned}t &= \sqrt{\frac{2\, h}{g}} && (\text{$t \ge 0$)} \end{aligned}[/tex].
(So is the case for other free fall motions where the initial vertical velocity is [tex]0[/tex].)
Substitute in [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex] and the initial height of the baseball [tex]h = 3\; \rm m[/tex]:
[tex]\begin{aligned}t &= \sqrt{\frac{2\, h}{g}} \\ &= \sqrt{\frac{3\; \rm m}{9.81\; \rm m\cdot s^{-2}}} \\ &\approx 0.782\; \rm s\end{aligned}[/tex].
In other words, the baseball would have flown for approximately [tex]0.782\; \rm s[/tex] before landing. If there is no air resistance on this baseball, the horizontal velocity of this baseball would be constant ([tex]50\; \rm m\cdot s^{-1}[/tex] until the ball lands.) This baseball would have travelled a horizontal distance of approximately:
[tex]50\; \rm m \cdot s^{-1} \times 0.782\; \rm s \approx 39\; \rm m[/tex].
