the acceleration of the gravity on the moon is 1.62m/s^2. if a ball is dropped on the moon from a height of 1.50m, determine the time for the ball to fall the surface of the moon
h = 1/2 a t^2 ... 1.50 = 1/2 * 1.62 * t^2
solve for t
To determine the time it takes for the ball to fall to the surface of the moon, we can use the equation:
h = (1/2) * g * t^2
where:
h = height (1.50 m)
g = acceleration due to gravity (1.62 m/s^2)
t = time
First, rearrange the equation to solve for time (t):
t^2 = (2h) / g
t = sqrt((2h) / g)
Substitute the given values into the equation:
t = sqrt((2 * 1.50 m) / 1.62 m/s^2)
t = sqrt(3 / 1.62)
t ≈ sqrt(1.85)
t ≈ 1.36 seconds
Therefore, the time it takes for the ball to fall to the surface of the moon is approximately 1.36 seconds.
To determine the time for the ball to fall to the surface of the moon, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
s = displacement (1.50m in this case, as the ball is dropped from a height of 1.50m)
u = initial velocity (0 m/s, as the ball is dropped, not thrown or given an initial velocity)
a = acceleration due to gravity on the moon (1.62 m/s^2)
t = time taken for the ball to fall
Rearranging the equation to solve for t, we have:
s = (1/2)at^2
2s = at^2
t^2 = (2s)/a
t = √((2s)/a)
Now let's substitute the given values:
s = 1.50 m
a = 1.62 m/s^2
t = √((2 * 1.50) / 1.62)
Calculating further:
t = √(3 / 1.62)
t = √(3 / 1.62)
t ≈ √1.85
t ≈ 1.36 seconds
Therefore, it takes approximately 1.36 seconds for the ball to fall to the surface of the moon.