This can not be so hard a problem.
1、The rock first moves upward until it reachs the highest point ,then fall down.
It's initial antrorse speed of the rock is 30m/s while it's acceleration is just the acceleration of the gravity.
According to the equation V2^2 - V1^2 = 2as
while the V2 = 0 ,V1=30m/s and a = -g = -10m/s^2
So ,the hight of the rock moving up is s = 30*30/(2*10) = 45m
2、As we know ,the whole progress can be divided into 2 parts.
At first ,the rock run up until its speed is 0 .
The time is t1 ,so t1 = V2-V1/(-g) = 3s .
Then the rock act free-fall motion while the hight is 325 + 45 = 370m.According to the equation s = 0.5gt^2 ,we assume that the time is t2 ,so t2 = √（2*370/10） = 8.6s .
Finally the whole time is T = t1 + t2 = 11.6s
3、If it were launched staright down ,the rock has an downward initial speed of 30m/s while the acceleration of gravity acts.
According to the equation s = V1*t + 0.5gt^2
while s = 325m ,V1 = 30m/s and g = 10m/s^2
So the key to the equation is the solve of the proble.
t = √（74）- 3 = 5.6s
4、To be honest ,I can not sure the meaning of the word “dropped”.
Because there is no angle we known ,so I can only assume that the initial speed is horizontal.
So the time of the rock running in the air depend on the time of motion of the vertical free-fall motion.
The time is t and the equation is s = 0.5gt^2
while s = 325m ,g = 10m/s^2
so t = √65 = 8.06s