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.