To determine the time it takes for a ball thrown vertically upward with an initial speed of 20 m/s to return to the thrower's hand, we need to understand the motion of the ball in the context of the physics of free fall under gravity.
### Step-by-Step Solution:
1. Initial Information and Assumptions:
- Initial speed of the ball, [tex]\( u = 20 \, \text{m/s} \)[/tex].
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] (standard gravity).
- When the ball returns to the thrower's hand, its displacement, [tex]\( s = 0 \)[/tex].
2. Time to Reach the Maximum Height:
- When the ball reaches its maximum height, its final velocity, [tex]\( v \)[/tex], is 0 m/s.
- The time to reach the maximum height, [tex]\( t_{\text{up}} \)[/tex], can be calculated using the formula for uniformly accelerated motion:
[tex]\[
v = u - g \cdot t_{\text{up}}
\][/tex]
- Setting [tex]\( v = 0 \)[/tex] and solving for [tex]\( t_{\text{up}} \)[/tex]:
[tex]\[
0 = 20 - 9.8 \cdot t_{\text{up}}
\][/tex]
[tex]\[
9.8 \cdot t_{\text{up}} = 20
\][/tex]
[tex]\[
t_{\text{up}} = \frac{20}{9.8} \approx 2.0408163265306123 \, \text{seconds}
\][/tex]
3. Total Time for the Ball to Return to the Thrower:
- The total time for the ball to go up and come back down is twice the time taken to reach the maximum height:
[tex]\[
t_{\text{total}} = 2 \cdot t_{\text{up}}
\][/tex]
[tex]\[
t_{\text{total}} = 2 \cdot 2.0408163265306123 \approx 4.081632653061225 \, \text{seconds}
\][/tex]
Thus, the time taken for the ball to return to the thrower's hand is approximately 4.08 seconds.
