an astronaut standing on a platform on the moon drops a hammer. if the hammer falls 6.0 meters vertically in 2.7 seconds, what os its acceleration.
d=1/2 a t^2
solve for a.
To find the acceleration of the hammer, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Since the hammer is dropping vertically, we can assume the initial velocity is 0 m/s (as the hammer is initially at rest). The final velocity can be determined using the distance it falls and the time it takes to fall, using the formula:
distance = (1/2) * acceleration * time^2
We know the distance (6.0 meters) and time (2.7 seconds), and we can rearrange the formula to solve for acceleration:
acceleration = (2 * distance) / (time^2)
Now we can substitute the given values to calculate the acceleration:
acceleration = (2 * 6.0 meters) / (2.7 seconds)^2
Simplifying the equation, we have:
acceleration = 1.185 m/s^2
Therefore, the acceleration of the hammer is approximately 1.185 m/s^2.
To determine the acceleration of the hammer, we can use the basic kinematic equation:
\[ d = \frac{1}{2} \cdot a \cdot t^2 \]
where:
- \( d \) is the distance (vertical displacement) traveled by the hammer (6.0 meters in this case),
- \( a \) is the acceleration, and
- \( t \) is the time taken (2.7 seconds in this case).
Rearranging the equation to solve for acceleration (\( a \)), we get:
\[ a = \frac{2d}{t^2} \]
Substituting the given values:
\[ a = \frac{2 \cdot 6.0}{(2.7)^2} \]
Now, let's calculate the acceleration:
\[ a = \frac{12.0}{7.29} \]
\[ a \approx 1.646 \, \text{m/s}^2 \]
Therefore, the hammer's acceleration is approximately \( 1.646 \, \text{m/s}^2 \).