The acceleration of gravity on the moon is 1.62 m/s² if a ball is dropped on the moon from a height of 1.50m determine the time for the ball to fall to the surface of the moon
h = 1/2 g t^2 ... t = √(2 h / g) = √(2 * 1.50 / 1.62) ... seconds
To determine the time for the ball to fall to the surface of the moon, we can use the equation:
distance = (1/2) * acceleration * time^2,
where distance is the vertical distance the ball travels, acceleration is the acceleration due to gravity on the moon (1.62 m/s²), and time is the time it takes for the ball to reach the surface.
In this case, the distance is given as 1.50 meters. Let's solve the equation to find the time.
Rearranging the equation, we have:
time^2 = (2 * distance) / acceleration
Plugging in the values, we get:
time^2 = (2 * 1.50) / 1.62
time^2 = 3.00 / 1.62
time^2 = 1.852
Taking the square root of both sides, we find:
time ≈ √1.852
time ≈ 1.36 seconds
So, the time 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 equation of motion:
s = ut + 0.5at^2
Where:
s = height (1.50 m)
u = initial velocity (0 m/s since the ball is dropped)
a = acceleration due to gravity on the moon (1.62 m/s²)
t = time
First, let's rearrange the equation to solve for time (t):
s = ut + 0.5at^2
1.50 = 0t + 0.5 * 1.62 * t^2
1.50 = 0.81t^2
Divide both sides of the equation by 0.81:
1.50 / 0.81 = t^2
1.85 = t^2
Take the square root of both sides to solve for t:
t = √1.85
t ≈ 1.36 seconds
Therefore, it will take approximately 1.36 seconds for the ball to fall to the surface of the moon.