A rock falls from rest a vertical distance of 0.63 meters to the surface of a planet in 0.48 seconds. What is the magnitude of the acceleration due to gravity on that planet?
d = Vo*t + 0.5g*t^2 = 0.63m,
0 + 0.5g*(0.48)^2 = 0.63,
0.1152g = 0.63,
g = 5.47m/s^2.
To find the magnitude of the acceleration due to gravity on the planet, we can use the equation for free fall motion:
Δy = 1/2 * g * t^2
Where:
Δy = vertical distance
g = acceleration due to gravity
t = time
In this case, the vertical distance is given as 0.63 meters and the time taken is 0.48 seconds. Plugging these values into the equation, we have:
0.63 = 1/2 * g * (0.48)^2
Now we can solve for g by rearranging the equation:
g = (2 * 0.63) / (0.48)^2
g = 2.625
Therefore, the magnitude of the acceleration due to gravity on that planet is approximately 2.625 m/s^2.
To find the magnitude of the acceleration due to gravity on the planet, we can use the equation:
d = (1/2) * a * t^2,
where:
d is the vertical distance traveled (0.63 meters),
a is the acceleration due to gravity, and
t is the time taken (0.48 seconds).
Rearranging the equation, we get:
a = (2 * d) / t^2.
Substituting the given values, we have:
a = (2 * 0.63) / 0.48^2.
Simplifying, we find:
a ≈ 8.24 m/s^2.
Therefore, the magnitude of the acceleration due to gravity on that planet is approximately 8.24 m/s^2.