If a ball is thrown in the air with an initial height of 4 feet, and if the ball remains in the air for 3.8 seconds, then accurate to the nearest foot, how high did it go? Remember, the acceleration due to gravity on Earth is \( -32 \mathrm{ft} / \mathrm{sec}^{2} \). [?] feet

1. Write the equation for vertical motion: \[ y(t) = y_0 + v_0 t - 16t^2, \] where \( y_0 = 4 \) ft, \( v_0 \) is the initial velocity, and the acceleration due to gravity is \( -32 \) ft/sec\(^2\) (hence the term \(-16t^2\) because \( \frac{1}{2}(-32) = -16 \)). 2. Since the ball remains in the air for \( 3.8 \) seconds and lands when \( y(3.8) = 0 \), substitute \( t = 3.8 \) into the equation: \[ 0 = 4 + v_0(3.8) - 16(3.8)^2. \] 3. Compute \( 16(3.8)^2 \): \[ 3.8^2 = 14.44 \quad \text{and} \quad 16 \times 14.44 = 231.04. \] Thus, the equation becomes: \[ 0 = 4 + 3.8\,v_0 - 231.04. \] 4. Solve for \( v_0 \): \[ 3.8\,v_0 = 231.04 - 4 = 227.04, \] \[ v_0 = \frac{227.04}{3.8} \approx 59.74 \, \text{ft/sec}. \] 5. The maximum height is reached when the velocity becomes zero. The velocity at time \( t \) is given by: \[ v(t) = v_0 - 32t. \] Set \( v(t) = 0 \) to find the time to reach the maximum height: \[ 0 = 59.74 - 32t_{\text{max}}, \] \[ t_{\text{max}} = \frac{59.74}{32} \approx 1.867 \, \text{sec}. \] 6. Substitute \( t_{\text{max}} \) into the position equation to find the maximum height: \[ y_{\text{max}} = 4 + 59.74(1.867) - 16(1.867)^2. \] First, calculate: \[ 59.74 \times 1.867 \approx 111.45, \] and \[ 1.867^2 \approx 3.4849 \quad \text{so} \quad 16 \times 3.4849 \approx 55.76. \] Hence, \[ y_{\text{max}} \approx 4 + 111.45 - 55.76 = 59.69 \, \text{ft}. \] 7. Rounding to the nearest foot: \[ y_{\text{max}} \approx 60 \, \text{ft}. \]