A ball was thrown straight up at 20.7 m/s then caught at the same place it was released. How many seconds was it in the air?
To find out how many seconds the ball was in the air, we can use the formula for free-falling objects:
\[ d = v_i \cdot t + \frac{1}{2} \cdot g \cdot t^2 \]
Where:
- \( d \) is the displacement (change in height) of the ball
- \( v_i \) is the initial velocity of the ball
- \( t \) is the time the ball is in the air
- \( g \) is the acceleration due to gravity, approximately 9.8 m/s^2
In this case, the ball was thrown straight up, so the initial velocity (\( v_i \)) is positive (20.7 m/s), and the displacement (\( d \)) is zero since the ball was caught at the same place it was released.
Substituting these values into the formula, we have:
\[ 0 = 20.7 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
Now we can solve for \( t \). Rearranging the equation, we get:
\[ 9.8 \cdot t^2 + 20.7 \cdot t = 0 \]
Factoring out \( t \), we have:
\[ t(9.8t + 20.7) = 0 \]
This equation is satisfied if either \( t = 0 \) or \( 9.8t + 20.7 = 0 \). However, the time cannot be zero (as it would mean the ball didn't move), so we solve the second equation:
\[ 9.8t = -20.7 \]
Dividing both sides by 9.8, we find:
\[ t = \frac{-20.7}{9.8} \]
Evaluating this expression, we get:
\[ t \approx -2.112 \, \text{s} \]
Since time cannot be negative, we discard the negative sign. Therefore, the ball was in the air for approximately 2.112 seconds.