A feather is dropped onto the surface of the moon. How far will the feather have fallen if it reaches the surface in 9.00 seconds?
Given: g on moon = -1.63 meters/second2
h = Vo*t + 0.5g*t^2.
h = 0 + 0.5*1.63*(9)^2 = 66 m.
a feather is dropped from the surface of the moon. how far will the feather have fallen if it reaches the surface in 9.00 seconds?
To solve this problem, we can use the equation for the distance fallen by an object under constant acceleration:
d = (1/2) * g * t^2
where:
d = distance fallen
g = acceleration due to gravity
t = time
In this case, we are given that the acceleration due to gravity on the moon, g, is -1.63 meters/second^2, and the time taken, t, is 9.00 seconds.
Using the given values, we can substitute them into the equation:
d = (1/2) * (-1.63) * (9.00)^2
Calculating the equation:
d = (1/2) * (-1.63) * 81.00
d = -0.815 * 81.00
d = -66.12 meters
Therefore, if the feather reaches the surface of the moon in 9.00 seconds, it will have fallen approximately 66.12 meters.
To calculate the distance the feather will have fallen on the moon in 9.00 seconds, we can use the equation of motion:
d = (1/2) * g * t^2
Where:
d is the distance fallen
g is the acceleration due to gravity on the moon (-1.63 m/s^2)
t is the time taken (9.00 s)
Now, let's substitute the values into the equation and solve for d:
d = (1/2) * (-1.63 m/s^2) * (9.00 s)^2
First, let's square the time:
d = (1/2) * (-1.63 m/s^2) * (81.00 s^2)
Next, let's multiply the terms:
d = (-0.815 m/s^2) * (81.00 s^2)
d = -66.135 m
Since distance cannot be negative, we can discard the negative sign. Therefore, the feather would have fallen approximately 66.135 meters on the surface of the moon in 9.00 seconds.