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# Confusing Physics Problem?

An object is released from rest on a planet that has no atmosphere. The object falls freely for 5.51 m in the first second.

What is the magnitude of the acceleration due to gravity on the planet?

Answer in units of m/s2

I thought it would be 5.51 m/s2, but apparently that's wrong. Does anyone know why and how to get the real answer?

### 2 Answers

- L. E. GantLv 79 years agoFavorite Answer
Look up the kinematic equations.

Find the one that has what you want to find and what you know.

In this case, it's "h = s + ut + at^2/2"

where h = distance travelled (=5.51metres), s = starting distance (=0), u = starting velocity (=0), a = acceleration (what you want to find) and t = time elapsed (=1 second)

So:

5.51 = 0 + 0 + a/2

a = 11.02 ((metres per second)/second)) or m/s^2

(The other guy didn't do the doubling right, but shows the same method)

- Anonymous9 years ago
s=5.51 u=0 a=a t=1

Use the equation s=ut+.5at^2

5.51=0t+.5a(1)^2

5.51=.5a

a=10.02 ms^-2