The rock has a height y at time t according to
y = 25 m + (60 m/s) sin(32°) t - 1/2 gt ²
where g = 9.8 m/s² is the magnitude of the acceleration due to gravity.
Solve for t when y = 0. The quickest way to do this is with the quadratic formula.
0 = 25 m + (60 m/s) sin(32°) t - 1/2 gt ²
t = (-(60 m/s) sin(32°) + √[(60 m/s)² sin²(32°) - 4 (25 m) (-1/2 gt ²)]) / 2
(ignoring the negative root because that would make t negative)
or approximately
t ≈ 7.2 s
