Question

On the Moon, acceleration resulting from gravity, g, is about 5.3 ft/s2. Which expression gives the time, in seconds, it would take a dropped penny to fall 100 ft on the Moon?

Answer 1

Answer:

$t=\sqrt{\frac{2h_0}{g}}$, 6.1 s

Explanation:

The motion of the dropped penny is a uniformly accelerated motion, with constant acceleration

$g=5.3 ft/s^2$

towards the ground. If the penny is dropped from a height of

$h_0 = 100 ft$

the vertical position of the penny at time t is given by the equation

$h(t) = h_0 - \frac{1}{2}gt^2$

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

$0=h_0 - \frac{1}{2}gt^2$

And re-arranging it, we find an expression for the time t:

$t=\sqrt{\frac{2h_0}{g}}$

And substituting the numbers, we can also find the numerical value:

$t=\sqrt{\frac{2(100 ft)}{5.3 ft/s^2}}=6.1 s$

Answer 2

Answer:

$20\sqrt[]{5/53}$