Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is
v = 30 m/s - g t
where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by
y = (30 m/s) t - 1/2 g t ²
(a) At its maximum height, the particle has 0 velocity, which occurs for
0 = 30 m/s - g t
t = (30 m/s) / g ≈ 3.06 s
at which point the particle's maximum height would be
y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m
(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.
(c) The particle falls 35 m below its starting point when
-35 m = (30 m/s) t - 1/2 g t ²
Solve for t to get a time of about t ≈ 7.1 s