Let us choose the origin, which coincides with the initial position of the ball. Then the equations of vertical motion are "y(t) = v_0 t - \\frac{g t^2}{2}", "v(t) = v_0 - g t".
At highest point, the speed is zero, hence "v(t') = v_0 - g t'", from where the time to reach that point is "t' = \\frac{v_0}{g}". Hence, the ball rose at height "y(t') = \\frac{v_0^2}{g} - \\frac{g}{2} \\frac{v_0^2}{g^2} = \\frac{v_0^2}{2 g} \\approx 11.47 m".
To find the time the ball remains in the air, equate its y coordinate to zero, and solve for time:
"y(t) = v_0 t - \\frac{g t^2}{2} = t(v_0 - \\frac{g t}{2}) = 0", from where "t = 0" (initially the ball is at the origin) or "t = \\frac{2 v_0}{g} \\approx 3.1 s". Hence, the ball remains "3.1 s" in the air.
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