Abaseball dropped from the roof of a tall building takes 5.23 second s to reach the ground. how tall is the building?

The most general formula useful here would be

s = s0 + v0*t + (1/2)at^2, where s0 is the original height, v0 is the original velocity, a is the acceleration, and t is the time.

Here, s becomes 0 (baseball is on the ground)
           v0 is 0, since the initial speed of the ball was 0.
            a is the acc'n due to gravity, and is -9.8 m per (sec)^2 in the metric sys.
            t is known and is 5.23 sec

All that is left to find is s0, the original height (which is the building height).

0 = s0 + (0)(5.23 sec) + (1/2) (-9.8 m per sec^2)(5.23 sec)^2

 -s0     =         0             - (4.9 m per sec^2) ( 27.35 sec^2)

 So               s0 = 134 meters (approx)

The building height was approx. 134 meters.

The answer of the above is

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