To find the distance the ball is lifted, we can use the formula for work done, which is given by:
\[ W = F \cdot d \cdot \cos(\theta) \]
where:
- \( W \) is the work done (in joules),
- \( F \) is the net force (in newtons),
- \( d \) is the distance (in meters),
- \( \theta \) is the angle between the force and the direction of motion.
In this case, since the ball is being lifted straight upwards, the angle \( \theta = 0^\circ \) and \( \cos(0^\circ) = 1\). Therefore, the equation simplifies to:
\[ W = F \cdot d \]
Rearranging this to solve for \( d \):
\[ d = \frac{W}{F} \]
Now substitute the given values into the equation:
\[ W = 4.35 \times 10^{-2} , \text{J} \] \[ F = 7.25 \times 10^{-2} , \text{N} \]
Now plug in these values:
\[ d = \frac{4.35 \times 10^{-2}}{7.25 \times 10^{-2}} \]
Calculating this gives:
\[ d = \frac{4.35}{7.25} \times \frac{10^{-2}}{10^{-2}} = \frac{4.35}{7.25} \approx 0.6 \]
Therefore, the distance the ball is lifted is approximately:
\[ d \approx 0.60 , \text{m} \]
This means the ball is lifted approximately 0.60 meters.