Answer:
28.7 m.
Explanation:
Hello,
In this case, since the coin is dropped from the rest condition at which the initial velocity is 0 m/s, we can compute the height of the precipice via the following equation:
[tex]y=y_0+v_0*t-\frac{1}{2}gt^2[/tex]
Whereas the final height is 0, the initial one is to be computed and the gravity is considered as 9.8 m/s², therefore the height turns out:
[tex]y=y_0+v_0*t-\frac{1}{2}gt^2\\\\y_0=\frac{1}{2}gt^2=\frac{1}{2}*9.8m/s^2*(2.42s)^2 \\\\y_0=28.7m[/tex]
It means that the height of the cliff is 28.7 m.
Regards.
