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:
[tex]v'^{2} = v^{2} + 2gh[/tex]
[tex]v'^{2} = 11.8^{2} + 2\times 9.8\times 34.0 = 805.64[/tex]
[tex]v' = \sqrt{805.64} = 28.38 m/s[/tex]
Now, time taken by the coin to reach the ground is given by eqn (1):
v' = v + gt
[tex]t = \frac{v' - v}{g} = \frac{28.38 - 11.8}{9.8} = 1.69 s[/tex]
