Calculated the total time kinematics

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A stone is thrown upward from a 70 m cliff with an initial speed of 12.0 m/s. The calculated time to reach the bottom is 5.2 seconds. To find the speed just before impact and the total distance traveled, it's suggested to determine the maximum height first, where the velocity is zero. The final speed before hitting the ground is calculated to be 38.9 m/s, and the total distance traveled is 84.7 m. Understanding the kinematic equations and the concept of maximum height is crucial for solving the remaining parts of the problem.
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Hi,
Here is the problem that am working on.

A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 70 m high.
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting
(c) What total distance did it travel.

I have calculated the total time it takes to reach the bottom of the cliff, but haven't been able to do the (b) and (c) part. Here is what i have done

I used this formula to compute the time and am getting it right.

x= x(0) + v(0)t + 1/2 at^2

0= 70+12t+1/2(-9.8)t^2
t= 5.2 seconds, -2.75 seconds

the ans to the c part is 84.7m and b part is38.9 m/s. Please suggest me a hint. Thanks
 
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Looks like you need to find the maximum height the stone makes it to, for B) and C). Remember at the max height velocity = 0
 
And remember that the V0 will be positive in the velocity equation
 
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