To solve this problem, let's break it down into a few steps.
### Step-by-Step Solution:
1. Determine the initial speed of the ball with respect to the ground:
- The lift is moving upwards with a speed of [tex]\(10 \, \text{m/s}\)[/tex].
- The boy throws the ball upwards with a speed of [tex]\(24.5 \, \text{m/s}\)[/tex] relative to the lift.
- Therefore, the total initial speed of the ball with respect to the ground is the sum of these two speeds:
[tex]\[
\text{Initial speed of the ball} = 10 \, \text{m/s} + 24.5 \, \text{m/s} = 34.5 \, \text{m/s}
\][/tex]
2. Calculate the time it takes for the ball to reach its highest point:
- At the highest point, the speed of the ball will be [tex]\(0 \, \text{m/s}\)[/tex].
- We use the equation of motion [tex]\( v = u - g t \)[/tex], where [tex]\(v\)[/tex] is the final speed (0 m/s at the highest point), [tex]\(u\)[/tex] is the initial speed (34.5 m/s), [tex]\(g\)[/tex] is the acceleration due to gravity (10 m/s²), and [tex]\(t\)[/tex] is the time taken to reach the highest point.
- Rearranging the formula, we get:
[tex]\[
t = \frac{u}{g} = \frac{34.5 \, \text{m/s}}{10 \, \text{m/s}^2} = 3.45 \, \text{s}
\][/tex]
This is the time taken to reach the highest point.
3. Calculate the total time for the ball to return to the boy's hand:
- The total time for the ball to return to the boy's hand is twice the time taken to reach the highest point, as the time to go up is equal to the time to come down.
[tex]\[
\text{Total time} = 2 \times \text{time to highest point} = 2 \times 3.45 \, \text{s} = 6.9 \, \text{s}
\][/tex]
### Final Result:
Thus, the ball will return to the boy’s hand in [tex]\(6.9\)[/tex] seconds.
