Answer:
To find the speed of the ball 0.70 seconds after it is released, we can use the equations of motion under constant acceleration. The acceleration due to gravity on the moon is \( 1.62 \, \text{m/s}^2 \).
Given:
- Initial velocity (\( u \)) of the ball: \( 0.50 \, \text{m/s} \) (downward)
- Initial height (\( h \)) of the ball: \( 4.0 \, \text{m} \)
- Acceleration due to gravity (\( g \)) on the moon: \( 1.62 \, \text{m/s}^2 \)
We can use the equation of motion:
\[ v = u + at \]
Where:
- \( v \) is the final velocity
- \( u \) is the initial velocity
- \( a \) is the acceleration
- \( t \) is the time
Substituting the values, we have:
\[ v = 0.50 \, \text{m/s} + (1.62 \, \text{m/s}^2)(0.70 \, \text{s}) \]
\[ v = 0.50 \, \text{m/s} + 1.134 \, \text{m/s} \]
\[ v = 1.634 \, \text{m/s} \]
Therefore, the speed of the ball 0.70 seconds after it is released is \( 1.634 \, \text{m/s} \) downward.
