If the object had an initial launch speed of 10.0 m/s and was launched at 30 degrees to the horizontal,

Initial y velocity = 10sin(30) = 5 m/s

Initial x velocity = 10cos(30) = 8.66 m/s

Time it takes to get to the top of the flight: vfy = viy + at, where a = -9.81 m/s^2, viy = 5 m/s, and vfy = 0 m/s (Initial poin t = launch, Final point = top of path). You should find that t = 0.51 seconds.

The flight path is perfectly symmetrical, so the total time of flight is 0.51 seconds * 2 = 1.02 seconds.

Flight range is x velocity multiplied by time: Range = vx * t = 8.66 m/s * 1.02 s = 8.83 meters.

Max height can be calculated using 2ay = vfy^2 - viy^2, where a = -9.81 m/s^2, vfy = 0 m/s, and viy = 5 m/s (The initial point is the precise moment of launch and the final point is the top of the flight when vfy = 0 m/s). The calculated height is y = 1.27 meters.

So we found that range is 8.83 m and height is 1.27 m. 8.83/1.27 = 6.92 or so. If you didn't round, the answer becomes 7. This could be done using only variables as well, but having numbers to compare is nice.