The time t (in seconds) that it takes for an object to fall a distance of s feet is given by the formula

t = radical symbol s /4

In some places, the Grand Canyon is one mile (5,280 feet) deep. How long would it take a stone dropped over the edge of the canyon to hit bottom?

would the answer be

18 or 26 seconds?

d = (1/2)g t^2

in feet, seconds units g = 32 ft/s^2 approximately
so
d = 16 t^2
t^2 = d/16
t = (1/4)(sqrt d)
if d = 5280
t = .25 (sqrt 5280)
= .25 * 72.7
=18.2 seconds

I get neither of those. Put this into the google search window:

sqrt(5280/4)

Finally, your formula is not accurate.

ok i got

36.3318042

thank you

would the formula be

t = sqrt of 5280/4?

for a sense check - check in SI units!

1 mile is approx 1600 metres, a is approx 10 m s^-2

t=sqrt(2s/a)

t = sqrt(2 x 1600/10)

t= sqrt(320)

t is approx 18 (8x sqrt5)

or have I made a mistake?

Jennifer

I answered this same question for you yesterday.

http://www.jiskha.com/display.cgi?id=1224095175

To find the time it takes for a stone to fall a distance of 5,280 feet in the Grand Canyon, you can use the given formula:

t = √(s/4),

where t is the time in seconds and s is the distance in feet.

Substituting the given distance of 5,280 feet into the formula, we have:

t = √(5280/4).

Simplifying the expression inside the square root:

t = √(1320).

Taking the square root of 1320:

t ≈ 36.33 seconds.

Therefore, it would take approximately 36.33 seconds for a stone to hit the bottom of the Grand Canyon when dropped from its edge.