Time of flight, t = 5 seconds.
From equations of motion,
y = vot - 1/2gt^2
v = vo -gt
At maximum height, v = 0 and t = 2.5 seconds
Therefore,
0 = vo -9.81*2.5 => vo = 9.81*2.5 = 24.525 m/s
Therefore,
f(x) = 24.525t - 1/2gt^2 = 24.525t - 1/2*9.81t = 24.525t - 4.905t^2
In other words, practical domain of f(x) is;
f(x) = 24.525t - 4.905t&2