Answer:
a. The rock's velocity is
and the acceleration is 
b. It takes 22.5 seconds to reach the highest point.
c. The rock goes up to 405 m.
d. It reach half its maximum height when time is 6.59 s or 38.41 s.
e. The rock is aloft for 45 seconds.
Step-by-step explanation:
- Velocity is defined as the rate of change of position or the rate of displacement.

- Acceleration is defined as the rate of change of velocity.

a.
The rock's velocity is the derivative of the height function 

The rock's acceleration is the derivative of the velocity function 

b. The rock will reach its highest point when the velocity becomes zero.

It takes 22.5 seconds to reach the highest point.
c. The rock reach its highest point when t = 22.5 s
Thus

So the rock goes up to 405 m.
d. The maximum height is 405 m. So the half of its maximum height = 
To find the time it reach half its maximum height, we need to solve

For a quadratic equation of the form
the solutions are


It reach half its maximum height when time is 6.59 s or 38.41 s.
e. It is aloft until s(t) = 0 again

The rock is aloft for 45 seconds.