The easiest way to figure this out is to draw a parabola and put marks of equal height on the left and right sides to visualize it. The bottom left is t = o and the bottom right is the Tfinal. Mark 2.7 on the left side and 5.3 on the right side (at equal heights). If you move up both sides a little, you'll find that a mark on the left at 3 seconds will be at the same height as a mark at 5 seconds on the right side. This means the top must be at 4 seconds.
If that doesn't work, then consider this. At 2.7 seconds, it must be traveling up and at 5.3 seconds it must be traveling down. Given that it is at the same height, it must have 2.7 seconds of fall time still left at its 5.3-second height. That makes the total flight time 5.3 + 2.7 = 8 seconds. If it starts at 0 and finishes at 8 seconds, then it's at its highest point at 4 seconds.
Lastly you could simply average 2.7 and 5.3 to find that it was halfway between those two times at 4 seconds, and halfway through a parabola is the top.
From there the easiest thing to do is use the BR shortcut where vfinal = 10t in freefall, so from the top until the ground (which takes 4 seconds total) you reach an impact speed of roughly 40 m/s. Because of the symmetry of the parabolic flight, if the final speed is 40 m/s, then the launch speed must also be 40 m/s.
