A ball is thrown straight up into the air at \( 18.2 \mathrm{~m} / \mathrm{s} \). How far has the ball traveled after 1.00 s ?

To find the distance the ball has traveled after 1.00 s, we can use the kinematic equation for the vertical motion of an object under constant acceleration due to gravity: \[ s = ut + \frac{1}{2}at^2 \] where: - \( s \) is the distance traveled, - \( u \) is the initial velocity, - \( a \) is the acceleration (in this case, the acceleration due to gravity, which is approximately \( -9.8 \, \text{m/s}^2 \) in the downward direction), - \( t \) is the time. Given: - Initial velocity, \( u = 18.2 \, \text{m/s} \) - Time, \( t = 1.00 \, \text{s} \) - Acceleration due to gravity, \( a = -9.8 \, \text{m/s}^2 \) Substitute the values into the equation and calculate the distance traveled by the ball after 1.00 s. Calculate the value by following steps: - step0: Calculate: \(18.2\times 1+0.5\left(-9.8\right)\times 1^{2}\) - step1: Evaluate the power: \(18.2\times 1+0.5\left(-9.8\right)\times 1\) - step2: Multiply: \(18.2+0.5\left(-9.8\right)\times 1\) - step3: Multiply the terms: \(18.2-4.9\) - step4: Subtract the numbers: \(13.3\) The ball has traveled a distance of \( 13.3 \, \text{m} \) after 1.00 s.