# Cliff Free fall question

The equation for free fall at the surface of the planet Quixon is s = 3.8t2m with t in sec. Assume a rock is dropped from the top of a 400-m cliff. Find the speed of the rock at t = 6 sec.
The problem is.. i don't know if the "m" is part of the equation or not.. like should i plug in the 400 as
s = 3.8t2(400) and then find the derivative and plug in 6 for t..
or should i just interpret the m as "meters" because s is a function to find position.. which would be in meters..
and in this case i'd just find the derivative of 3.8t2 and then plug in 6 for t...
hmm?

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Jacobpm64 said:
should i just interpret the m as "meters" because s is a function to find position.. which would be in meters..
and in this case i'd just find the derivative of 3.8t2 and then plug in 6 for t...
hmm?
I think that's what you should do.

then the 400 would be just extra information that is useless?

Jacobpm64 said:
then the 400 would be just extra information that is useless?
Well no it would be the initial height of the projectile, so if you were asked the height of the projectile after a time t it would be

400 - 3.8$$t^2$$

all right, but i wasn't asked the height :P.. but is the answer 45.6 m/s?

Jacobpm64 said:
all right, but i wasn't asked the height :P.. but is the answer 45.6 m/s?
Depending on their definition of speed it's either that or half that. If they mean speed as in distance traveled divided by time traveled then its 22.8, but if they mean speed as in absolute value of the instantaneous velociy then you're right.

HallsofIvy