An object is released from rest on a planet
that has no atmosphere. The object falls
freely for 5.03 m in the first second.
What is the magnitude of the acceleration
due to gravity on the planet?
Answer in units of m/s2
10.06
To find the magnitude of the acceleration due to gravity on the planet, we can use the formula for free fall motion:
h = (1/2) * g * t^2
where:
h = distance fallen (5.03 m)
g = acceleration due to gravity
t = time (1 second)
Plugging in the values, we can rearrange the equation to solve for g:
g = (2 * h) / t^2
g = (2 * 5.03 m) / (1 s)^2
g = 10.06 m/s^2
Therefore, the magnitude of the acceleration due to gravity on the planet is 10.06 m/s^2.
To calculate the magnitude of the acceleration due to gravity on the planet, we can use the formula for free fall motion:
d = (1/2) * g * t^2
where:
d = distance traveled (5.03 m in this case)
g = acceleration due to gravity
t = time (1 second in this case)
We can rearrange the formula to solve for g:
g = (2 * d) / t^2
Plugging in the values, the equation becomes:
g = (2 * 5.03 m) / (1 s)^2
Simplifying further:
g = 10.06 m/s^2
Therefore, the magnitude of the acceleration due to gravity on the planet is 10.06 m/s^2.