The acceleration of gravity on the moon is 1.62 m/s2. If a ball is dropped on
the moon from a height of 1.50 m. Determine the time for the ball to fall to the surface of the moon.
1.36s
the acceleration of gravity on the moon is 1.62m/s2. if a ball is dropped on the moon from a height of 1.50m. determine the time for the ball to fall to the surface of the moon
0.45s
distance = 1/2 * gt^2
You know d and g, substitute and solve for t.
Well, on the moon, gravity plays a little game of hide and seek with the objects. In this case, the acceleration due to gravity on the moon is 1.62 m/s^2, which is approximately one-sixth of the acceleration due to gravity on Earth. So, let's calculate the time for our little ball's moonwalk!
We can use the kinematic equation that relates the distance, acceleration, and time. In this case, the distance is 1.50 m, the acceleration is 1.62 m/s^2, and we're solving for the time. Plugging in these values, we get:
distance = (0.5) * acceleration * time^2
1.50 m = 0.5 * 1.62 m/s^2 * time^2
Now, let's solve for time and see how long our ball will take to reach the lunar surface!
time^2 = (2 * 1.50 m) / 1.62 m/s^2
time^2 = 2.77 seconds
Taking the square root of both sides, we get:
time = √(2.77 seconds)
And the moment you've all been waiting for...
time ≈ 1.663 seconds!
So, according to my calculations, it will take approximately 1.663 seconds for our ball to fall to the surface of the moon. Just enough time for it to practice some moon acrobatics!
To determine the time it takes for the ball to fall to the surface of the moon, we can use the equation for motion under constant acceleration:
h = (1/2) * g * t^2
Where:
h = height (1.50 m)
g = acceleration due to gravity (1.62 m/s^2)
t = time
We can rearrange the equation to solve for time:
t^2 = (2h) / g
t = sqrt[(2h) / g]
Now, let's substitute the given values:
t = sqrt[(2 * 1.50 m) / 1.62 m/s^2]
t = sqrt[3.00 / 1.62]
t ≈ sqrt(1.85)
t ≈ 1.36 seconds
Therefore, it takes approximately 1.36 seconds for the ball to fall to the surface of the moon.