Answer:
The ball would hit the floor approximately [tex]0.55\; \rm s[/tex] after leaving the table.
The ball would travel approximately [tex]5.5\; \rm m[/tex] horizontally after leaving the table.
(Assumption: [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].)
Explanation:
Let [tex]\Delta h[/tex] denote the change to the height of the ball. Let [tex]t[/tex] denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let [tex]v_0(\text{vertical})[/tex] denote the initial vertical velocity of this ball.
If the air resistance on this ball is indeed negligible:[tex]\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t[/tex].
The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was [tex]v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}[/tex].
The height of the table was [tex]1.5\; \rm m[/tex]. Therefore, after hitting the floor, the ball would be [tex]1.5\; \rm m \![/tex] below where it was before leaving the table. Hence, [tex]\Delta h = -1.5\;\rm m[/tex].
The equation becomes:
[tex]\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}[/tex].
Solve for [tex]t[/tex]:
[tex]\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55[/tex].
In other words, it would take approximately [tex]0.55\; \rm s[/tex] for the ball to hit the floor after leaving the table.
Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at [tex]v(\text{horizontal}) =10\; \rm m \cdot s^{-1}[/tex]) until the ball hits the floor.
The ball was in the air for approximately [tex]t = 0.55\; \rm s[/tex] and would have travelled approximately [tex]v(\text{horizontal})\cdot t \approx 5.5\;\rm m[/tex] horizontally during the flight.
