An outfielder throws a baseball with an initial speed of \(32 \mathrm{~m} / \mathrm{s}\) at an angle of \(23^{\circ}\) to the horizontal. The ball leaves his hand from a height of \(1.83 \mathrm{~m}\). How long is the ball in the air before it hits the ground?

First, we need to determine the initial vertical velocity of the baseball. To do this, we will use the given initial speed and angle to find the vertical component of the velocity using the sine function. The equation for the initial vertical velocity is: \(v_{y0} = v_0 \sin(\theta)\) \(v_{y0} = 32 \mathrm{~m/s} \sin(23^{\circ})\) \(v_{y0} ≈ (32)(0.3907) = 12.5 \mathrm{~m/s}\)

Next, we need to find the time it takes for the baseball to reach the highest point. To do this, we will use the equation of motion for a projectile in the vertical direction: \(v_y = v_{y0} - g t\) Here, \(v_y\) is the final vertical velocity at the highest point (which is \(0\)), \(v_{y0}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)), and \(t\) is the time. We will solve for the time when \(v_y = 0\). \(0 = 12.5 \mathrm{~m/s} - 9.81t\) \(9.81t = 12.5 \mathrm{~m/s}\) \(t = \frac{12.5 \mathrm{~m/s}}{9.81 \mathrm{~m/s^2}}\) \(t ≈ 1.27 \mathrm{~s}\)

Now, we need to find the total time the baseball is in the air. We will use the following equation of motion for a projectile in the vertical direction: \(y = y_0 + v_{y0} t - \frac{1}{2} g t^2\) Here, \(y\) is the final vertical position (which is \(0\)), \(y_0\) is the initial vertical position, \(v_{y0}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time. We will solve for the time when \(y = 0\) and \(y_0 = 1.83 \mathrm{~m}\). \(0 = 1.83 \mathrm{~m} + 12.5 \mathrm{~m/s} t - \frac{1}{2}(9.81 \mathrm{~m/s^2})(t^2)\) This is a quadratic equation, so we will use the quadratic formula to solve for \(t\): \(t = \frac{-v_{y0} \pm \sqrt{(v_{y0}^2 - 2gy_0)}}{g}\) Using the calculated value of \(v_{y0}\), we get: \(t = \frac{-12.5 \mathrm{~m/s} \pm \sqrt{(12.5^2 - 2(9.81)(1.83))}}{9.81}\) \(t ≈ -0.149 \mathrm{~s}\) or \(t ≈ 2.55 \mathrm{~s}\) Since the time cannot be negative, we choose the positive value \(t = 2.55 \mathrm{~s}\). The ball is in the air for approximately \(2.55\) seconds before it hits the ground.