Calculus question?

Question

When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 4 feet above the lunar surface. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the 4 feet. How long did it take the hammer and feather to fall 4 feet on the moon, if the moon's gravitational acceleration is 13/80 that of Earth?

Answer 1

Nice question, but it uses such standard kinematical results, that I wouldn't call it a "calculus question":

On earth, accn. due to gravity, g_E is 32 ft sec^(-2). *** So on the Moon, the question tells us that:

g_M = 13 x 32 / 80 or 26 / 5 ft sec^(-2)

Distance travelled from rest with accn. 'a' is (a t^2 / 2).

So, 4 = (26 / 5) t^2 / 2. Therefore t^2 = 40 / 26 = 20 / 13 sec^2.

Using a calculator, t = 1.24 secs approximately.

So it would take the feather and hammer ~ 1.24 seconds, or about 1 1/4 seconds, to fall from a height of 4 feet on the Moon.

(*** Note that more significant figures are not justified, because of the limited accuracy of the input data. Also, I used g_E = 32, rather than 32.2 ft sec^(-2), consistently with your use of "only half a second to fall the 4 feet" on Earth. That latter result is exact for a "32" g_E measure and 4 feet, but not for the more accurate "32.2" g_E measure and 4 feet.)

Live long and prosper.

Answer 2

i could say shape is a factor of engineering meaning that a sturdy hold close on math is important, pre-cal is is like algebra to the subsequent point with the Pi chart and understanding approximately radians and perspective measures (which seems important) yet Calculus is the be taught of derivatives, (the slope of a line at a definite element) form of pointless no be counted what field of workd your going to, in case you question me.

Answer 3

First of all, I don't think we ever landed on the moon...but to answer your question:

s = 1/2 at^2

a of earth = 9.8 m/s^2 or 32.2 ft/sec^2
a of moon = 13/80 (32.2) = 5.23 ft/sec^2

s = 4 feet

t = (2s/a)^.5 = (2*4/5.23)^.5= 1.24 sec

Answer 4

13/80 * 32.2 = 5.2325 ft/s/s

t = sqrt(2h/g) = sqrt(2*4/5.2325)

t = sqrt(1.5289) = 1.2365 seconds (on moon)

t = sqrt(2*4/32.2) = .3332 seconds on earth