Answer:
[tex]t=\sqrt{\frac{2h_0}{g}}[/tex], 6.1 s
Explanation:
The motion of the dropped penny is a uniformly accelerated motion, with constant acceleration
[tex]g=5.3 ft/s^2[/tex]
towards the ground. If the penny is dropped from a height of
[tex]h_0 = 100 ft[/tex]
the vertical position of the penny at time t is given by the equation
[tex]h(t) = h_0 - \frac{1}{2}gt^2[/tex]
where the negative sign is due to the fact that the direction of the acceleration is downward.
We want to know the time t at which the penny reaches the ground, which means h(t)=0. Substituting into the equation, it becomes
[tex]0=h_0 - \frac{1}{2}gt^2[/tex]
And re-arranging it, we find an expression for the time t:
[tex]t=\sqrt{\frac{2h_0}{g}}[/tex]
And substituting the numbers, we can also find the numerical value:
[tex]t=\sqrt{\frac{2(100 ft)}{5.3 ft/s^2}}=6.1 s[/tex]
