The acceleration of gravity on the moon is 1.62m/s.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.
To determine the time for the ball to fall to the surface of the moon, we can use the equation for motion under constant acceleration:
\[ d = v_i t + \frac{1}{2} a t^2 \]
where:
- \( d \) is the distance traveled (in this case, the height),
- \( v_i \) is the initial velocity (which is zero),
- \( a \) is the acceleration (1.62 m/s² on the moon),
- \( t \) is the time it takes for the ball to fall.
Plugging in the given values, we have:
\[ 1.50 = 0 \cdot t + \frac{1}{2} \cdot 1.62 \cdot t^2 \]
Simplifying the equation gives:
\[ 1.50 = 0.81 \cdot t^2 \]
Now, let's solve for \( t \):
\[ t^2 = \frac{1.50}{0.81} \]
\[ t^2 = 1.852 \]
\[ t = \sqrt{1.852} \]
Calculating the square root, we find that:
\[ t \approx 1.36 \, \text{seconds} \]
Therefore, it takes approximately 1.36 seconds for the ball to fall to the surface of the moon.
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 = distance or height (1.50 m)
u = initial velocity (0 m/s, as the ball is dropped)
a = acceleration (acceleration due to gravity on the moon = 1.62 m/s^2)
t = time
Rearranging the equation, we have:
(1/2)at^2 = s
Substituting the given values, we get:
(1/2)(1.62 m/s^2)t^2 = 1.50 m
Simplifying further:
0.81t^2 = 1.50 m
Now, solve for t by isolating t:
t^2 = 1.50 m / 0.81 m/s^2
t^2 = 1.85 s^2
Taking the square root of both sides, we find:
t ≈ √(1.85) s
t ≈ 1.36 s
So, the time for the ball to fall to the surface of the moon is approximately 1.36 seconds.
so, 45 minutes ago you posted this same question and said it was answered.
Why are you posting it again? Give it a rest.