An astronaut drops a golf ball to the surface of the moon from a height of 1.5m. The golf ball has a velocity of 2.24 m/s right before it lands on the moon (assume this height to be zero). What is acceleration due to gravity on the moon?
g= vĀ²/2h
t = 1.36 s
To determine the acceleration due to gravity on the moon, we can use the equation of motion for an object in free fall:
h = ut + (1/2)gt^2
where:
h = height
u = initial velocity
g = acceleration due to gravity
t = time
In this case, the height (h) is 1.5m, the initial velocity (u) is 2.24 m/s, and the time (t) is the time taken for the golf ball to fall to the surface of the moon.
Since the height at the end is 0, we can rewrite the equation as:
0 = (1/2)gt^2
Simplifying, we get:
gt^2 = 0
Since t cannot be 0 (as the ball had to fall for some amount of time), the only possibility is that g is equal to 0. Hence, the acceleration due to gravity on the moon is 0 m/s^2.
This result may seem counterintuitive, but it is because the moon has a weaker gravitational field compared to Earth.
To find the acceleration due to gravity on the moon, we can use the equations of motion. The key equation we'll use is:
v^2 = u^2 + 2as
Where:
v = final velocity (2.24 m/s)
u = initial velocity (0 m/s, as the ball was dropped, not thrown)
a = acceleration (which is the acceleration due to gravity on the moon)
s = displacement (1.5 m, as the ball falls from a height of 1.5 m)
We can rearrange the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Plugging in the given values:
a = (2.24^2 - 0^2) / (2 * 1.5)
Simplifying the equation:
a = 5.0176 / 3
a ā 1.672 m/s^2
Therefore, the acceleration due to gravity on the moon is approximately 1.672 m/s^2.