Answer:
Approximately [tex]6\; {\rm s}[/tex], assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance is negligible.
Explanation:
When the projectile was launched from ground level and travels upward, speed decreases while height increases, and kinetic energy ([tex]\text{KE}[/tex]) is converted into gravitational potential energy ([tex]\text{GPE}[/tex]). As the projectile returns to the ground, the opposite happens- speed increases while height decreases, and [tex]\text{GPE}[/tex] is converted back into [tex]\text{KE}[/tex].
Under the assumption that air resistance is negligible, sum of [tex]\text{GPE}[/tex] and [tex]\text{KE}[/tex] would stay the same during the entire flight. Since the projectile returned to ground (same height,) [tex]\text{GPE}[/tex] at landing would be the same as [tex]\text{GPE}[/tex] at launch. Since the sum of [tex]\text{GPE}[/tex] and [tex]\text{KE}[/tex] need to stay unchanged, [tex]\text{KE}[/tex] at landing would also need to be the same as the value at launch, and the speed of the projectile at landing should be the same as the speed at launch: [tex]30\; {\rm m\cdot s^{-1}}[/tex].
However, note that despite that speed (magnitude of velocity) has returned to the original value, the direction of velocity of the projectile at landing would be opposite to the direction at launch. If the velocity at launch is [tex]u = 30\; {\rm m\cdot s^{-1}}[/tex], the velocity at landing would be [tex](-30)\; {\rm m\cdot s^{-1}}[/tex], negative because the projectile would be travelling downward.
Also under the assumption that air resistance is negligible, the projectile would accelerate towards the ground at a constant [tex]a = -g = (-9.81)\; {\rm m\cdot s^{-2}}[/tex], negative because this acceleration points downwards. To find the duration of the motion, divide the change in velocity [tex](v - u)[/tex] by acceleration:
[tex]\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{(-30\; {\rm m\cdot s^{-1}}) - (30\; {\rm m\cdot s^{-1}})}{(-9.81\; {\rm m\cdot s^{-1}})} \\ &\approx 6\; {\rm s}\end{aligned}[/tex].
In other words, it would take approximately [tex]6[/tex] seconds for this projectile to return to the ground.