Answer:
Time taken by the coin to reach the ground is 1.69 s
Given:
Initial speed, v = 11.8 m/s
Height of the building, h = 34.0 m
Solution:
Now, from the third eqn of motion:
![v'^(2) = v^(2) + 2gh](https://img.qammunity.org/2020/formulas/physics/college/ehjn2mkdgkd0o0luiii1g8pybykqiziftu.png)
![v'^(2) = 11.8^(2) + 2* 9.8* 34.0 = 805.64](https://img.qammunity.org/2020/formulas/physics/college/4sv4hpxsf9jc8kdndf1sfzjmiqjpggh9yt.png)
![v' = √(805.64) = 28.38 m/s](https://img.qammunity.org/2020/formulas/physics/college/r2diu9lz63dn367zpobcdtj7fb7p7jxsf3.png)
Now, time taken by the coin to reach the ground is given by eqn (1):
v' = v + gt
![t = (v' - v)/(g) = (28.38 - 11.8)/(9.8) = 1.69 s](https://img.qammunity.org/2020/formulas/physics/college/i0389q00ayf8jk76bpk2hvp383dvuv1p3a.png)