If a ball is thrown in the air with an initial height of 4 feet, and if the ball remains in the air for 4.5 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

To find the maximum height that the ball reaches, we can use the physics of projectile motion. The upward motion can be modeled by the equation \( h(t) = -16t^2 + v_0 t + h_0 \), where \( h(t) \) is the height at time \( t \), \( v_0 \) is the initial velocity, and \( h_0 \) is the initial height. First, we need to determine the initial velocity \( v_0 \). Since the ball is thrown from a height of 4 feet and it impacts the ground after 4.5 seconds, we set \( h(4.5) = 0 \). Plugging in the values, we rearrange the equation to solve for \( v_0 \): \[ 0 = -16(4.5)^2 + v_0(4.5) + 4 \] This simplifies to: \[ 0 = -324 + 4.5v_0 + 4 \] So, \[ 4.5v_0 = 320 \] Thus, \[ v_0 \approx 71.1 \text{ ft/s} \] Now that we have \( v_0 \), we can find the time at which the ball reaches its maximum height, which occurs when the velocity is zero. The velocity function is given by: \[ v(t) = v_0 - 32t \] Setting \( v(t) = 0 \): \[ 0 = 71.1 - 32t \] Solving for \( t \): \[ t \approx 2.22 \text{ seconds} \] Now, substituting \( t \) back into the height equation: \[ h(2.22) = -16(2.22)^2 + 71.1(2.22) + 4 \] Calculating this gives us: \[ h(2.22) \approx 80 \text{ feet} \] Thus, the ball reaches a maximum height of approximately 80 feet.