acceleration due to gravity questions
when an object is dropped from a height of 10m above the surface of the planet z, it takes 1.2 seconds for the object to reach the surface. what is the acceleration of a falling object near the surface of the planet?
Distance fallen = (1/2) a t^2
Solve for a
is it 1.6m/s^2
No. Did you use
a = 2 (Height)/t^2 ?
To find the acceleration of a falling object near the surface of a planet, we can use the equation:
h = (1/2) * g * t^2
where:
- h is the height from which the object is dropped (10m in this case),
- g is the acceleration due to gravity,
- t is the time taken for the object to reach the surface (1.2s in this case).
Rearranging the equation to solve for g, we have:
g = (2 * h) / (t^2)
Now we can substitute the given values into the equation:
g = (2 * 10m) / (1.2s)^2
Simplifying further:
g = (2 * 10m) / 1.44s^2
≈ 13.89 m/s^2
Therefore, the acceleration of the falling object near the surface of planet Z is approximately 13.89 m/s^2.