A crewman throws a rock

[miscellaneous]

A crewman on the starship Enterprise is on shore leave on a distant planet. He drops a rock from the top of a cliff and observes that it takes 3.00 s to reach the bottom. He now throws another rock vertically upwards so that it reaches a height of 2.00 m before dropping down the cliff face. The second rock takes 4.12 s to reach the bottom of the cliff. The planet has a very thin atmosphere that offers negligible air resistance. (a) How high is the cliff? (b) What is the value of g, the acceleration due to gravity, on the planet?

Here's how I did this:

2 + h = whole height which took 4.12 seconds

2+h = .5a(4.12)^2
h= 8.4872a - 2

h= .5a(3)^2
h= 4.5a

8.4872a-2 = 4.5a
a= .5 m/s^2

h=4.5a
h= 4.5(.5)
h= 2.257 m

Can anyone PLEASE verify these answers? Thanks a lottt

[Doc Al]

He now throws another rock vertically upwards so that it reaches a height of 2.00 m before dropping down the cliff face. The second rock takes 4.12 s to reach the bottom of the cliff.
I assume that the time of 4.12 s is the total time from when the rock is throw to when it hits the ground, not merely the time it took to fall from its maximum height to the ground.

[miscellaneous]

I think that if that was the case, the problem would be virtually impossible without more information because we don't know at what acceleration the rock was thrown vertically and therefore couldn't tell the velocity-->we couldn't tell that time..Do you agree? If I'm wrong, please let me know.

[Doc Al]

I think you have all the info needed to solve for the height of the cliff and the acceleration.

[miscellaneous]

Can you please guide me in the right direction? I dont know what I should do..

[Doc Al]

Start by writing every equation that fits the data:
(1) rock thrown rises to a height of 2 m
(2) rock thrown hits the ground in 4.12 s
(3) rock dropped hits the ground in 3 s