Answer:
Approximately [tex]34\; {\rm m\cdot s^{-1}}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Assume that the drag on the dart is negligible. Vertically, this dart will be accelerating downward at [tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex] under the gravitational pull.
Since this dart was launched horizontally, the initial vertical velocity of this dart will be [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex].
The height of the dart has changed by [tex]0.6\; {\rm m}[/tex]. Thus, the vertical displacement of this dart will be [tex]x = 0.6\; {\rm m}[/tex].
Let [tex]t[/tex] denote the amount of time the dart spent in the air. Since the acceleration is constant in the vertical component, the SUVAT equation [tex]x = (1/2)\, a\, t^{2} + u\, t[/tex] would apply in that component:
[tex]x = (1/2)\, a\, t^{2} + u\, t[/tex].
[tex](0.6) = (1/2)\, (9.81)\, t^{2} + (0) \, t[/tex].
[tex](0.6) = (1/2)\, (9.81)\, t^{2}[/tex].
Note that the term [tex]u\, t[/tex] is eliminated since the initial vertical velocity of this dart is [tex]0\; {\rm m\cdot s^{-1}}[/tex] as the dart was launched horizontally.
Rearrange and solve this equation for [tex]t[/tex]:
[tex]\begin{aligned} (0.6) = \frac{1}{2}\, (9.81)\, t^{2}\end{aligned}[/tex].
[tex]\begin{aligned} \frac{1}{2}\, (9.81)\, t^{2} = (0.6)\end{aligned}[/tex].
[tex]\begin{aligned} (9.81)\, t^{2} = (2)\, (0.6)\end{aligned}[/tex].
[tex]\begin{aligned} t^{2} = \frac{(2)\, (0.6)}{(9.81)}\end{aligned}[/tex].
[tex]\begin{aligned} t &= \sqrt{\frac{(2)\, (0.6)}{(9.81)}} \\ &\approx 0.349\end{aligned}[/tex].
Thus, it would have taken this dart approximately [tex]0.349\; {\rm s}[/tex] to travel from where it was launched to where it landed. Since the dart has travelled a horizontal distance of [tex]12\; {\rm m}[/tex] in that amount of time, the initial horizontal velocity of this dart would be:
[tex]\begin{aligned}(\text{horizontal velocity}) &= \frac{(\text{horizontal displacement})}{(\text{time taken})} \\ &\approx \frac{12\; {\rm m}}{0.349\; {\rm s}} \\&\approx 34\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].
