Calculated the total time kinematics

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The discussion focuses on a kinematics problem involving a stone thrown vertically upward with an initial speed of 12.0 m/s from a height of 70 m. The calculated time to reach the bottom of the cliff is 5.2 seconds, using the equation x = x(0) + v(0)t + 1/2 at^2. The final speed of the stone just before impact is determined to be 38.9 m/s, and the total distance traveled is 84.7 m. Key insights include the importance of finding the maximum height to solve for the final speed and distance.

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jacy
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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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